YES
We show the termination of the TRS R:

  p(|0|()) -> |0|()
  p(s(x)) -> x
  plus(x,|0|()) -> x
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  plus(s(x),y) -> s(plus(p(s(x)),y))
  plus(x,s(y)) -> s(plus(x,p(s(y))))
  times(|0|(),y) -> |0|()
  times(s(|0|()),y) -> y
  times(s(x),y) -> plus(y,times(x,y))
  div(|0|(),y) -> |0|()
  div(x,y) -> quot(x,y,y)
  quot(zero(y),s(y),z) -> |0|()
  quot(s(x),s(y),z) -> quot(x,y,z)
  quot(x,|0|(),s(z)) -> s(div(x,s(z)))
  div(div(x,y),z) -> div(x,times(zero(y),z))
  eq(|0|(),|0|()) -> true()
  eq(s(x),|0|()) -> false()
  eq(|0|(),s(y)) -> false()
  eq(s(x),s(y)) -> eq(x,y)
  divides(y,x) -> eq(x,times(div(x,y),y))
  prime(s(s(x))) -> pr(s(s(x)),s(x))
  pr(x,s(|0|())) -> true()
  pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
  if(true(),x,y) -> false()
  if(false(),x,y) -> pr(x,y)
  zero(div(x,x)) -> x
  zero(divides(x,x)) -> x
  zero(times(x,x)) -> x
  zero(quot(x,x,x)) -> x
  zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(s(x),y) -> plus#(x,y)
p2: plus#(s(x),y) -> plus#(p(s(x)),y)
p3: plus#(s(x),y) -> p#(s(x))
p4: plus#(x,s(y)) -> plus#(x,p(s(y)))
p5: plus#(x,s(y)) -> p#(s(y))
p6: times#(s(x),y) -> plus#(y,times(x,y))
p7: times#(s(x),y) -> times#(x,y)
p8: div#(x,y) -> quot#(x,y,y)
p9: quot#(s(x),s(y),z) -> quot#(x,y,z)
p10: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p11: div#(div(x,y),z) -> div#(x,times(zero(y),z))
p12: div#(div(x,y),z) -> times#(zero(y),z)
p13: div#(div(x,y),z) -> zero#(y)
p14: eq#(s(x),s(y)) -> eq#(x,y)
p15: divides#(y,x) -> eq#(x,times(div(x,y),y))
p16: divides#(y,x) -> times#(div(x,y),y)
p17: divides#(y,x) -> div#(x,y)
p18: prime#(s(s(x))) -> pr#(s(s(x)),s(x))
p19: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y))
p20: pr#(x,s(s(y))) -> divides#(s(s(y)),x)
p21: if#(false(),x,y) -> pr#(x,y)
p22: zero#(s(x)) -> if#(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))
p23: zero#(s(x)) -> eq#(x,s(|0|()))
p24: zero#(s(x)) -> plus#(zero(|0|()),|0|())
p25: zero#(s(x)) -> zero#(|0|())
p26: zero#(s(x)) -> plus#(|0|(),zero(|0|()))

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The estimated dependency graph contains the following SCCs:

  {p8, p9, p10, p11, p13, p17, p19, p20, p21, p22}
  {p7}
  {p1, p2, p4}
  {p14}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: pr#(x,s(s(y))) -> divides#(s(s(y)),x)
p2: divides#(y,x) -> div#(x,y)
p3: div#(div(x,y),z) -> zero#(y)
p4: zero#(s(x)) -> if#(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))
p5: if#(false(),x,y) -> pr#(x,y)
p6: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y))
p7: div#(div(x,y),z) -> div#(x,times(zero(y),z))
p8: div#(x,y) -> quot#(x,y,y)
p9: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p10: quot#(s(x),s(y),z) -> quot#(x,y,z)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r3, r4, r8, r11, r13, r17, r18, r19, r20, r23, r27

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        pr#_A(x1,x2) = max{219, x1 + 89, x2 + 111}
        s_A(x1) = x1 + 108
        divides#_A(x1,x2) = max{x1 + 109, x2 + 63}
        div#_A(x1,x2) = max{91, x1 + 62}
        div_A(x1,x2) = max{x1 + 132, x2 + 306}
        zero#_A(x1) = x1 + 306
        if#_A(x1,x2,x3) = max{x2 + 89, x3 + 219}
        eq_A(x1,x2) = max{x1 + 107, x2 - 11}
        |0|_A = 1
        plus_A(x1,x2) = max{x1 + 40, x2 + 19}
        zero_A(x1) = max{67, x1 - 131}
        false_A = 59
        divides_A(x1,x2) = max{171, x1 - 120, x2 + 1}
        times_A(x1,x2) = x1 + 41
        quot#_A(x1,x2,x3) = max{91, x1 + 62}
        quot_A(x1,x2,x3) = max{130, x1, x2, x3 - 109}
        true_A = 108
        pr_A(x1,x2) = max{170, x1 - 167, x2 - 168}
  
    base order 2:
  
      precedence:
      
        div = pr > s = eq > pr# = div# = zero# = if# = |0| = quot# > divides# = zero = divides = quot > plus = true > false = times
      
  
    partial status:
  
      pi(pr#) = [1]
      pi(s) = [1]
      pi(divides#) = []
      pi(div#) = [1]
      pi(div) = [1, 2]
      pi(zero#) = [1]
      pi(if#) = [2]
      pi(eq) = [1]
      pi(|0|) = []
      pi(plus) = [1]
      pi(zero) = []
      pi(false) = []
      pi(divides) = []
      pi(times) = []
      pi(quot#) = [1]
      pi(quot) = [1, 2]
      pi(true) = []
      pi(pr) = []

The following rules are removed:

  p1



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: divides#(y,x) -> div#(x,y)
p2: div#(div(x,y),z) -> zero#(y)
p3: zero#(s(x)) -> if#(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))
p4: if#(false(),x,y) -> pr#(x,y)
p5: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y))
p6: div#(div(x,y),z) -> div#(x,times(zero(y),z))
p7: div#(x,y) -> quot#(x,y,y)
p8: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p9: quot#(s(x),s(y),z) -> quot#(x,y,z)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The estimated dependency graph contains the following SCCs:

  {p6, p7, p8, p9}
  {p4, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p2: div#(x,y) -> quot#(x,y,y)
p3: quot#(s(x),s(y),z) -> quot#(x,y,z)
p4: div#(div(x,y),z) -> div#(x,times(zero(y),z))

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r1, r8, r11, r13, r17, r18, r19, r23, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2,x3) = x1 + 56
        |0|_A = 10
        s_A(x1) = max{21, x1 + 2}
        div#_A(x1,x2) = x1 + 56
        div_A(x1,x2) = max{31, x1 + 1, x2 - 24}
        times_A(x1,x2) = max{19, x1 - 46, x2 - 25}
        zero_A(x1) = max{21, x1 - 74}
        p_A(x1) = max{11, x1 - 38}
        quot_A(x1,x2,x3) = max{34, x1 - 22, x2 - 22, x3 - 24}
        eq_A(x1,x2) = max{20, x1 - 73, x2 - 22}
        true_A = 9
        false_A = 8
        pr_A(x1,x2) = max{25, x1 - 24, x2 - 13}
        if_A(x1,x2,x3) = max{21, x1 - 25, x2 - 25, x3 - 22}
  
    base order 2:
  
      precedence:
      
        s = div = quot = pr > times > quot# = div# > zero > eq = if > p > |0| > true > false
      
  
    partial status:
  
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []
      pi(div#) = []
      pi(div) = []
      pi(times) = []
      pi(zero) = []
      pi(p) = []
      pi(quot) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p4



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p2: div#(x,y) -> quot#(x,y,y)
p3: quot#(s(x),s(y),z) -> quot#(x,y,z)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p2: div#(x,y) -> quot#(x,y,y)
p3: quot#(s(x),s(y),z) -> quot#(x,y,z)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r1, r8, r11, r13, r17, r18, r19, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2,x3) = max{3, x1 + 1}
        |0|_A = 6
        s_A(x1) = x1 + 6
        div#_A(x1,x2) = max{3, x1 + 1}
        p_A(x1) = max{6, x1 - 42}
        times_A(x1,x2) = max{21, x1 - 21, x2 - 42}
        div_A(x1,x2) = max{20, x1 - 25, x2 - 47}
        quot_A(x1,x2,x3) = max{7, x1 - 44, x2 - 50, x3 - 50}
        zero_A(x1) = max{15, x1 - 82}
        eq_A(x1,x2) = max{14, x1 - 77, x2 - 22}
        true_A = 13
        false_A = 1
        if_A(x1,x2,x3) = max{5, x1 - 15, x2 - 39, x3 - 45}
  
    base order 2:
  
      precedence:
      
        times = div = true > s = quot = zero > eq = if > p > |0| > quot# = div# = false
      
  
    partial status:
  
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = [1]
      pi(div#) = [1]
      pi(p) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      pi(zero) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(if) = []

The following rules are removed:

  p3



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p2: div#(x,y) -> quot#(x,y,y)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p2: div#(x,y) -> quot#(x,y,y)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r1, r11, r13, r17, r18, r19, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2,x3) = max{4, x2 + 1, x3 - 1}
        |0|_A = 9
        s_A(x1) = 3
        div#_A(x1,x2) = x2 + 5
        p_A(x1) = max{12, x1 - 10}
        div_A(x1,x2) = max{6, x1 + 3, x2}
        quot_A(x1,x2,x3) = max{7, x1 + 3, x2 + 3, x3 + 3}
        zero_A(x1) = max{22, x1 + 13}
        eq_A(x1,x2) = max{0, x1 - 50, x2 - 6}
        true_A = 0
        false_A = 0
        if_A(x1,x2,x3) = max{22, x1, x2 - 49, x3 - 32}
  
    base order 2:
  
      precedence:
      
        p = div = true > quot# = |0| > s = div# = quot > zero > eq = if > false
      
  
    partial status:
  
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(div#) = []
      pi(p) = []
      pi(div) = [2]
      pi(quot) = [2, 3]
      pi(zero) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(if) = []

The following rules are removed:

  p1



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(x,y) -> quot#(x,y,y)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: if#(false(),x,y) -> pr#(x,y)
p2: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y))

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r1, r8, r13, r17, r18, r19, r20, r23, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if#_A(x1,x2,x3) = x3 + 3
        false_A = 1
        pr#_A(x1,x2) = max{2, x2 - 2}
        s_A(x1) = max{29, x1 + 6}
        divides_A(x1,x2) = max{34, x1 - 53, x2 - 40}
        p_A(x1) = max{81, x1 - 73}
        |0|_A = 32
        times_A(x1,x2) = max{34, x1, x2}
        quot_A(x1,x2,x3) = max{33, x1, x2, x3}
        zero_A(x1) = max{0, x1 - 149}
        eq_A(x1,x2) = max{42, x1, x2 - 39}
        true_A = 33
        pr_A(x1,x2) = max{35, x1 - 36, x2 - 37}
        if_A(x1,x2,x3) = max{5, x1 - 35, x3 - 37}
  
    base order 2:
  
      precedence:
      
        if# > pr# > s = p = quot = true > zero > |0| > false > if > pr > divides > times = eq
      
  
    partial status:
  
      pi(if#) = [3]
      pi(false) = []
      pi(pr#) = []
      pi(s) = [1]
      pi(divides) = []
      pi(p) = []
      pi(|0|) = []
      pi(times) = [1]
      pi(quot) = [1, 2, 3]
      pi(zero) = []
      pi(eq) = [1]
      pi(true) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p1



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y))

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: times#(s(x),y) -> times#(x,y)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r1, r8, r11, r13, r17, r18, r19, r23, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{x1 + 12, x2 + 13}
        s_A(x1) = x1 + 11
        p_A(x1) = max{38, x1 - 55}
        |0|_A = 37
        times_A(x1,x2) = max{40, x1 - 40, x2 - 111}
        div_A(x1,x2) = max{39, x1 - 70, x2 - 111}
        quot_A(x1,x2,x3) = max{38, x1 - 108, x2 - 145, x3 - 119}
        zero_A(x1) = max{10, x1 - 182}
        eq_A(x1,x2) = max{6, x1 - 172, x2 - 49}
        true_A = 5
        false_A = 1
        pr_A(x1,x2) = max{2, x1 - 62, x2 - 42}
        if_A(x1,x2,x3) = max{3, x1 - 42, x2 - 4, x3 - 23}
  
    base order 2:
  
      precedence:
      
        times > times# = p = div = true > s = quot > |0| > zero > eq = if > pr > false
      
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = [1]
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      pi(zero) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(s(x),y) -> plus#(x,y)
p2: plus#(x,s(y)) -> plus#(x,p(s(y)))
p3: plus#(s(x),y) -> plus#(p(s(x)),y)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r1, r2, r8, r11, r13, r17, r18, r19, r23, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1 + 2
        s_A(x1) = x1 + 31
        p_A(x1) = max{3, x1 - 30}
        |0|_A = 2
        times_A(x1,x2) = max{26, x1 - 32, x2 - 57}
        div_A(x1,x2) = max{25, x1 - 63, x2 - 90}
        quot_A(x1,x2,x3) = max{56, x1 - 32, x2 - 89, x3 - 58}
        zero_A(x1) = max{24, x1 - 154}
        eq_A(x1,x2) = max{8, x1 - 124, x2 - 34}
        true_A = 7
        false_A = 1
        pr_A(x1,x2) = max{8, x1 - 64, x2 - 136}
        if_A(x1,x2,x3) = max{23, x1 - 9, x2 - 28, x3 - 71}
  
    base order 2:
  
      precedence:
      
        times = quot > div = true > plus# = s > p = zero > |0| = eq = if > pr > false
      
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      pi(zero) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p1



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(x,s(y)) -> plus#(x,p(s(y)))
p2: plus#(s(x),y) -> plus#(p(s(x)),y)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(x,s(y)) -> plus#(x,p(s(y)))
p2: plus#(s(x),y) -> plus#(p(s(x)),y)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r2, r8, r11, r13, r17, r18, r19, r23, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1 + 11
        s_A(x1) = max{28, x1 + 24}
        p_A(x1) = max{14, x1 - 16}
        times_A(x1,x2) = max{56, x1 - 155, x2 - 176}
        |0|_A = 55
        div_A(x1,x2) = max{58, x1 - 59, x2 - 175}
        quot_A(x1,x2,x3) = max{83, x1 - 34, x2 - 203, x3 - 148}
        zero_A(x1) = max{26, x1 - 235}
        eq_A(x1,x2) = max{25, x1 - 212, x2 - 110}
        true_A = 20
        false_A = 21
        pr_A(x1,x2) = max{22, x1 - 49, x2 - 78}
        if_A(x1,x2,x3) = max{23, x1 - 58, x2 - 4, x3 - 63}
  
    base order 2:
  
      precedence:
      
        div > times = pr > if > plus# = s = |0| = quot > p = zero > eq > false > true
      
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(p) = []
      pi(times) = []
      pi(|0|) = []
      pi(div) = []
      pi(quot) = []
      pi(zero) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p2



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(x,s(y)) -> plus#(x,p(s(y)))

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(x,s(y)) -> plus#(x,p(s(y)))

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r1, r2, r8, r11, r13, r17, r18, r19, r23, r25, r30

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x2 + 1
        s_A(x1) = x1 + 9
        p_A(x1) = max{7, x1 - 7}
        |0|_A = 3
        times_A(x1,x2) = max{22, x1 - 17, x2 - 28}
        div_A(x1,x2) = max{14, x1 - 28, x2 - 51}
        quot_A(x1,x2,x3) = max{x1 - 14, x2 - 50, x3 + 81}
        zero_A(x1) = max{13, x1 - 80}
        eq_A(x1,x2) = max{9, x1 - 72, x2 - 23}
        true_A = 2
        false_A = 1
        pr_A(x1,x2) = max{3, x1 - 74, x2 - 65}
        if_A(x1,x2,x3) = max{4, x1 - 24, x2 - 1, x3 - 44}
  
    base order 2:
  
      precedence:
      
        times = quot > div > zero > s > plus# = eq = if > p > |0| > true > false > pr
      
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = [3]
      pi(zero) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: eq#(s(x),s(y)) -> eq#(x,y)

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: plus(x,|0|()) -> x
r4: plus(|0|(),y) -> y
r5: plus(s(x),y) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(p(s(x)),y))
r7: plus(x,s(y)) -> s(plus(x,p(s(y))))
r8: times(|0|(),y) -> |0|()
r9: times(s(|0|()),y) -> y
r10: times(s(x),y) -> plus(y,times(x,y))
r11: div(|0|(),y) -> |0|()
r12: div(x,y) -> quot(x,y,y)
r13: quot(zero(y),s(y),z) -> |0|()
r14: quot(s(x),s(y),z) -> quot(x,y,z)
r15: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r16: div(div(x,y),z) -> div(x,times(zero(y),z))
r17: eq(|0|(),|0|()) -> true()
r18: eq(s(x),|0|()) -> false()
r19: eq(|0|(),s(y)) -> false()
r20: eq(s(x),s(y)) -> eq(x,y)
r21: divides(y,x) -> eq(x,times(div(x,y),y))
r22: prime(s(s(x))) -> pr(s(s(x)),s(x))
r23: pr(x,s(|0|())) -> true()
r24: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r25: if(true(),x,y) -> false()
r26: if(false(),x,y) -> pr(x,y)
r27: zero(div(x,x)) -> x
r28: zero(divides(x,x)) -> x
r29: zero(times(x,x)) -> x
r30: zero(quot(x,x,x)) -> x
r31: zero(s(x)) -> if(eq(x,s(|0|())),plus(zero(|0|()),|0|()),s(plus(|0|(),zero(|0|()))))

The set of usable rules consists of

  r1, r8, r11, r13, r17, r18, r19, r23, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        eq#_A(x1,x2) = max{x1 - 1, x2 + 2}
        s_A(x1) = max{11, x1 + 3}
        p_A(x1) = max{0, x1 - 47}
        |0|_A = 0
        times_A(x1,x2) = max{0, x2 - 16}
        div_A(x1,x2) = max{48, x1 - 47, x2 - 47}
        quot_A(x1,x2,x3) = max{49, x1 - 43, x2 - 48, x3 - 48}
        zero_A(x1) = max{0, x1 - 47}
        eq_A(x1,x2) = max{0, x1 - 44, x2 - 95}
        true_A = 0
        false_A = 0
        pr_A(x1,x2) = max{0, x1 - 19, x2 - 21}
        if_A(x1,x2,x3) = max{0, x1 - 6, x2 - 19, x3 - 21}
  
    base order 2:
  
      precedence:
      
        times > eq# = div > s = quot > zero > eq = if > p = pr > |0| > true > false
      
  
    partial status:
  
      pi(eq#) = [2]
      pi(s) = [1]
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      pi(zero) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p1

Then no dependency pair remains.