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(|0|(),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(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)


-- 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(y,z))
p12: div#(div(x,y),z) -> times#(y,z)
p13: eq#(s(x),s(y)) -> eq#(x,y)
p14: divides#(y,x) -> eq#(x,times(div(x,y),y))
p15: divides#(y,x) -> times#(div(x,y),y)
p16: divides#(y,x) -> div#(x,y)
p17: prime#(s(s(x))) -> pr#(s(s(x)),s(x))
p18: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y))
p19: pr#(x,s(s(y))) -> divides#(s(s(y)),x)
p20: if#(false(),x,y) -> pr#(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(|0|(),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(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)

The estimated dependency graph contains the following SCCs:

  {p18, p20}
  {p8, p9, p10, p11}
  {p7}
  {p1, p2, p4}
  {p13}


-- 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(|0|(),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(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)

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:
      
        if#_A(x1,x2,x3) = x3 + 3
        false_A = 1
        pr#_A(x1,x2) = max{2, x2 - 28}
        s_A(x1) = max{52, x1 + 32}
        divides_A(x1,x2) = max{50, x1 - 59, x2 - 88}
        p_A(x1) = max{34, x1 - 120}
        |0|_A = 33
        times_A(x1,x2) = max{49, x1 - 53, x2 - 60}
        div_A(x1,x2) = max{48, x1 - 89, x2 - 139}
        quot_A(x1,x2,x3) = max{81, x1 - 56, x2 - 119, x3 - 106}
        eq_A(x1,x2) = max{33, x1 - 89, x2 - 53}
        true_A = 2
        pr_A(x1,x2) = max{51, x1 - 87, x2 - 54}
        if_A(x1,x2,x3) = max{85, x1 - 51, x2 - 86, x3 - 53}
  
    base order 2:
  
      precedence:
      
        if# = |0| = if > s = quot = true > p > pr# = pr > divides > times = eq > false = div
      
  
    partial status:
  
      pi(if#) = [3]
      pi(false) = []
      pi(pr#) = []
      pi(s) = []
      pi(divides) = []
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      pi(eq) = []
      pi(true) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: if#(false(),x,y) -> pr#(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(|0|(),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(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)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

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

The set of usable rules consists of

  r1, r3, r8, r11, 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:
      
        div#_A(x1,x2) = x1 + 39
        div_A(x1,x2) = max{4, x1 + 2, x2 - 28}
        times_A(x1,x2) = max{12, x1 + 11}
        quot#_A(x1,x2,x3) = x1 + 39
        |0|_A = 0
        s_A(x1) = x1 + 13
        p_A(x1) = x1 + 9
        plus_A(x1,x2) = x1 + 10
        quot_A(x1,x2,x3) = max{31, x1 + 18, x2, x3 - 14}
        eq_A(x1,x2) = max{x1 - 12, x2 + 12}
        true_A = 12
        false_A = 2
        pr_A(x1,x2) = max{13, x1 + 10, x2 - 4}
        if_A(x1,x2,x3) = max{13, x1 - 5, x2 - 1, x3 - 12}
  
    base order 2:
  
      precedence:
      
        plus > div = s = if > p = quot = eq > |0| = pr > div# = quot# = true > times = false
      
  
    partial status:
  
      pi(div#) = []
      pi(div) = []
      pi(times) = [1]
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(p) = [1]
      pi(plus) = [1]
      pi(quot) = [1, 2]
      pi(eq) = [2]
      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: div#(x,y) -> quot#(x,y,y)
p2: quot#(x,|0|(),s(z)) -> div#(x,s(z))
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(|0|(),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(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)

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

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:
      
        div#_A(x1,x2) = x1 + 2
        quot#_A(x1,x2,x3) = x1 + 2
        s_A(x1) = max{20, x1 + 3}
        |0|_A = 9
        p_A(x1) = max{18, x1 - 53}
        times_A(x1,x2) = max{22, x1 - 22, x2 - 29}
        div_A(x1,x2) = max{21, x1 - 29, x2 - 52}
        quot_A(x1,x2,x3) = max{25, x1 - 25, x2 - 53, x3 - 36}
        eq_A(x1,x2) = max{21, x1 - 29, x2 - 26}
        true_A = 8
        false_A = 4
        pr_A(x1,x2) = max{x1 + 24, x2 - 13}
        if_A(x1,x2,x3) = max{19, x1 - 26, x2 - 1, x3 - 21}
  
    base order 2:
  
      precedence:
      
        times > div = eq > div# = quot# = s = quot = pr = if > p > |0| > true > false
      
  
    partial status:
  
      pi(div#) = [1]
      pi(quot#) = [1]
      pi(s) = []
      pi(|0|) = []
      pi(p) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      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: div#(x,y) -> quot#(x,y,y)
p2: quot#(x,|0|(),s(z)) -> div#(x,s(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(|0|(),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(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)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

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:
      
        div#_A(x1,x2) = x2 + 3
        quot#_A(x1,x2,x3) = max{2, x2 + 1}
        |0|_A = 4
        s_A(x1) = 0
        p_A(x1) = x1
        times_A(x1,x2) = max{9, x1 - 10, x2 - 6}
        div_A(x1,x2) = max{6, x1 + 2, x2 + 2}
        quot_A(x1,x2,x3) = max{5, x1 + 2, x2 - 5, x3 + 2}
        eq_A(x1,x2) = max{x1 - 5, x2 + 2}
        true_A = 0
        false_A = 0
        pr_A(x1,x2) = max{0, x1 - 5, x2 - 2}
        if_A(x1,x2,x3) = max{x1 - 1, x2 - 1, x3 + 2}
  
    base order 2:
  
      precedence:
      
        times > p = div > quot# = |0| > div# = eq > s = quot = if > pr > true > false
      
  
    partial status:
  
      pi(div#) = [2]
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(p) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      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: quot#(x,|0|(),s(z)) -> div#(x,s(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(|0|(),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(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)

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(|0|(),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(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)

The set of usable rules consists of

  r1, r8, r11, 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:
      
        times#_A(x1,x2) = max{5, x1 - 14}
        s_A(x1) = max{20, x1 + 4}
        p_A(x1) = max{18, x1 - 115}
        |0|_A = 6
        times_A(x1,x2) = max{67, x1 - 69, x2 - 93}
        div_A(x1,x2) = max{68, x1 - 70, x2 - 208}
        quot_A(x1,x2,x3) = max{73, x1 - 21, x2 - 195, x3 - 193}
        eq_A(x1,x2) = max{21, x1 - 71, x2 - 68}
        true_A = 11
        false_A = 5
        pr_A(x1,x2) = max{12, x1 - 140, x2 - 150}
        if_A(x1,x2,x3) = max{25, x1 - 69, x2 - 139, x3 - 21}
  
    base order 2:
  
      precedence:
      
        times = eq > div = true = if > s = p = quot > |0| > times# = pr > false
      
  
    partial status:
  
      pi(times#) = []
      pi(s) = [1]
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      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(|0|(),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(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)

The set of usable rules consists of

  r1, r2, r8, r11, 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) = max{17, x1 + 4, x2 + 1}
        s_A(x1) = x1 + 13
        p_A(x1) = max{5, x1 - 12}
        |0|_A = 4
        times_A(x1,x2) = max{12, x1 - 21, x2 - 28}
        div_A(x1,x2) = max{19, x1 - 20, x2 - 33}
        eq_A(x1,x2) = max{14, x1 - 18, x2 - 14}
        true_A = 12
        false_A = 11
        pr_A(x1,x2) = max{26, x1 - 18, x2 - 56}
        if_A(x1,x2,x3) = max{27, x1 - 15, x2 - 1, x3 - 14}
  
    base order 2:
  
      precedence:
      
        times > plus# = div = eq = if > s > p > |0| > true > pr > false
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p3



-- SCC decomposition.

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)))

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(|0|(),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(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)

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#(s(x),y) -> plus#(x,y)
p2: 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(|0|(),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(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)

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:
      
        plus#_A(x1,x2) = x1 + 5
        s_A(x1) = x1 + 6
        p_A(x1) = max{8, x1 + 1}
        |0|_A = 4
        times_A(x1,x2) = max{13, x1 - 48, x2 - 15}
        div_A(x1,x2) = max{31, x1 - 33, x2 - 14}
        quot_A(x1,x2,x3) = max{37, x1 - 9, x2 - 13, x3 - 8}
        eq_A(x1,x2) = max{7, x1 - 33, x2 - 14}
        true_A = 2
        false_A = 1
        pr_A(x1,x2) = max{x1 + 33, x2 - 3}
        if_A(x1,x2,x3) = max{4, x1 - 30, x2 - 1, x3 - 7}
  
    base order 2:
  
      precedence:
      
        p = quot = pr > div = if > times > eq > plus# = s > |0| > true > false
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(p) = [1]
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      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)))

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(|0|(),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(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)

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(|0|(),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(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)

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) = max{6, x2 + 4}
        s_A(x1) = x1 + 20
        p_A(x1) = max{8, x1 - 15}
        |0|_A = 7
        times_A(x1,x2) = max{22, x1 - 22, x2 - 17}
        div_A(x1,x2) = max{21, x1 - 42, x2 - 65}
        quot_A(x1,x2,x3) = max{42, x1 - 21, x2 - 66, x3 - 21}
        eq_A(x1,x2) = max{21, x1 - 21, x2 - 23}
        true_A = 2
        false_A = 1
        pr_A(x1,x2) = max{3, x1 - 112, x2 - 92}
        if_A(x1,x2,x3) = max{2, x1 - 114, x3 - 93}
  
    base order 2:
  
      precedence:
      
        times > div = eq > s = p = quot > plus# = |0| > pr > true = if > false
      
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = []
      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(|0|(),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(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)

The set of usable rules consists of

  r1, r8, r11, 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:
      
        eq#_A(x1,x2) = max{x1 - 1, x2 + 4}
        s_A(x1) = max{17, x1 + 5}
        p_A(x1) = max{13, x1 - 6}
        |0|_A = 3
        times_A(x1,x2) = max{36, x1 - 21, x2 - 19}
        div_A(x1,x2) = max{4, x1 - 56, x2 - 149}
        quot_A(x1,x2,x3) = max{x1 - 50, x2 - 18, x3 + 19}
        eq_A(x1,x2) = max{18, x1 - 56, x2 - 37}
        true_A = 2
        false_A = 1
        pr_A(x1,x2) = max{3, x1 - 94, x2 - 84}
        if_A(x1,x2,x3) = max{23, x1 - 38, x2 - 1, x3 - 18}
  
    base order 2:
  
      precedence:
      
        times = quot > eq# = div = eq = if > s > p > |0| = pr > true > false
      
  
    partial status:
  
      pi(eq#) = [2]
      pi(s) = []
      pi(p) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = [3]
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p1

Then no dependency pair remains.