YES
We show the termination of the TRS R:

  plus(x,|0|()) -> x
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,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: times#(s(x),y) -> plus#(y,times(x,y))
p3: times#(s(x),y) -> times#(x,y)
p4: div#(x,y) -> quot#(x,y,y)
p5: quot#(s(x),s(y),z) -> quot#(x,y,z)
p6: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p7: div#(div(x,y),z) -> div#(x,times(y,z))
p8: div#(div(x,y),z) -> times#(y,z)
p9: eq#(s(x),s(y)) -> eq#(x,y)
p10: divides#(y,x) -> eq#(x,times(div(x,y),y))
p11: divides#(y,x) -> times#(div(x,y),y)
p12: divides#(y,x) -> div#(x,y)
p13: prime#(s(s(x))) -> pr#(s(s(x)),s(x))
p14: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y))
p15: pr#(x,s(s(y))) -> divides#(s(s(y)),x)
p16: if#(false(),x,y) -> pr#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p14, p16}
  {p4, p5, p6, p7}
  {p3}
  {p1}
  {p9}


-- 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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r1, r4, r5, r9, r13, r14, r15, r16, r21

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if#_A(x1,x2,x3) = max{3, x3 - 21}
        false_A = 3
        pr#_A(x1,x2) = max{2, x2 - 36}
        s_A(x1) = x1 + 20
        divides_A(x1,x2) = max{1, x1 - 105, x2 - 63}
        plus_A(x1,x2) = max{14, x1, x2 - 29}
        |0|_A = 27
        times_A(x1,x2) = max{x1 + 50, x2}
        quot_A(x1,x2,x3) = max{x1 - 93, x2 - 13, x3 + 28}
        eq_A(x1,x2) = max{4, x1 - 28, x2 - 21}
        true_A = 5
        if_A(x1,x2,x3) = max{41, x1 - 6, x2 - 61, x3 - 21}
  
    base order 2:
  
      precedence:
      
        plus = eq > divides > times = true = if > s = quot > pr# = |0| > if# = false
      
  
    partial status:
  
      pi(if#) = []
      pi(false) = []
      pi(pr#) = []
      pi(s) = [1]
      pi(divides) = []
      pi(plus) = [1]
      pi(|0|) = []
      pi(times) = [2]
      pi(quot) = [3]
      pi(eq) = []
      pi(true) = []
      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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: 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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r9, r13, r14, r15, r19, r21

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = max{x1 + 16, x2 + 9}
        div_A(x1,x2) = max{x1 + 3, x2 - 4}
        times_A(x1,x2) = max{10, x1 + 3, x2}
        quot#_A(x1,x2,x3) = max{x1 + 16, x2 + 9, x3 + 9}
        |0|_A = 5
        s_A(x1) = max{8, x1}
        plus_A(x1,x2) = max{x1, x2}
        quot_A(x1,x2,x3) = max{15, x1 + 7, x2 - 9, x3 - 9}
        eq_A(x1,x2) = max{9, x1 - 9, x2 - 11}
        true_A = 2
        false_A = 1
        pr_A(x1,x2) = max{8, x1 - 16, x2 - 10}
        if_A(x1,x2,x3) = max{9, x1 - 13, x2 - 1, x3 - 9}
  
    base order 2:
  
      precedence:
      
        div# = quot# > times > plus > s = if > div = quot = eq > |0| = pr > true > false
      
  
    partial status:
  
      pi(div#) = [1]
      pi(div) = [1]
      pi(times) = [1, 2]
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = [1]
      pi(plus) = [1, 2]
      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: 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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: 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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r4, r7, r9, r13, r14, r15, r19, r21

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = x1 + 10
        quot#_A(x1,x2,x3) = x1 + 10
        s_A(x1) = max{8, x1 + 2}
        |0|_A = 13
        times_A(x1,x2) = max{14, x1 - 71, x2 - 61}
        div_A(x1,x2) = max{27, x1 - 28, x2 - 71}
        quot_A(x1,x2,x3) = max{14, x1 - 62, x2 - 75, x3 - 72}
        eq_A(x1,x2) = max{4, x1 - 7, x2 - 15}
        true_A = 5
        false_A = 3
        pr_A(x1,x2) = max{6, x1 - 90, x2 - 90}
        if_A(x1,x2,x3) = max{11, x1 - 29, x2 - 89, x3 - 9}
  
    base order 2:
  
      precedence:
      
        |0| > div# = quot# = s = times = div = true = if > quot = eq > pr > false
      
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(s) = [1]
      pi(|0|) = []
      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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: 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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r13, r14, r15, r19, r21

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, x3 - 1}
        |0|_A = 15
        s_A(x1) = 10
        eq_A(x1,x2) = max{11, x1 - 10, x2 - 10}
        true_A = 7
        false_A = 6
        pr_A(x1,x2) = max{8, x1 - 20, x2 - 42}
        if_A(x1,x2,x3) = max{9, x1 - 16, x2 - 19, x3 - 31}
  
    base order 2:
  
      precedence:
      
        quot# = |0| = if > div# = s = true > eq > pr > false
      
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []
      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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: 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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r4, r7, r9, r13, r14, r15, r19, r21

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{4, x1 + 1}
        s_A(x1) = x1 + 8
        times_A(x1,x2) = max{10, x1 - 10, x2 - 18}
        |0|_A = 7
        div_A(x1,x2) = max{9, x1 - 24, x2 - 19}
        quot_A(x1,x2,x3) = max{18, x1 - 15, x2 - 22, x3 - 9}
        eq_A(x1,x2) = max{9, x1 - 14, x2 - 11}
        true_A = 3
        false_A = 6
        pr_A(x1,x2) = max{30, x1 + 4, x2 + 12}
        if_A(x1,x2,x3) = max{29, x1 - 27, x2 - 1, x3 - 7}
  
    base order 2:
  
      precedence:
      
        pr > times = eq = if > s = div = quot > times# = |0| > false > true
      
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = []
      pi(times) = []
      pi(|0|) = []
      pi(div) = []
      pi(quot) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = [2]
      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)

and R consists of:

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

The set of usable rules consists of

  r4, r7, r9, r13, r14, r15, r19, r21

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{7, x1 + 1}
        s_A(x1) = x1 + 14
        times_A(x1,x2) = max{16, x1 - 16, x2 - 31}
        |0|_A = 13
        div_A(x1,x2) = max{15, x1 - 30, x2 - 47}
        quot_A(x1,x2,x3) = max{30, x1 - 15, x2 - 46, x3 - 15}
        eq_A(x1,x2) = max{15, x1 - 20, x2 - 17}
        true_A = 10
        false_A = 12
        pr_A(x1,x2) = max{31, x1 + 7, x2 + 1}
        if_A(x1,x2,x3) = max{30, x1 + 1, x2 - 1, x3 - 9}
  
    base order 2:
  
      precedence:
      
        pr > times = eq = if > s = div = quot > plus# = |0| > false > true
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(times) = []
      pi(|0|) = []
      pi(div) = []
      pi(quot) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(pr) = [2]
      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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r4, r7, r9, r13, r14, r15, r16, r19, r21

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        eq#_A(x1,x2) = max{1, x1 - 17, x2 - 20}
        s_A(x1) = max{19, x1 + 10}
        times_A(x1,x2) = max{20, x1 - 37, x2 - 42}
        |0|_A = 7
        div_A(x1,x2) = max{8, x1 - 36, x2 - 43}
        quot_A(x1,x2,x3) = max{20, x1 - 25, x2 - 44, x3 - 32}
        eq_A(x1,x2) = max{20, x1 - 31, x2 + 11}
        true_A = 6
        false_A = 5
        pr_A(x1,x2) = max{17, x1 - 84, x2 - 74}
        if_A(x1,x2,x3) = max{30, x1 - 33, x2 - 1, x3 - 20}
  
    base order 2:
  
      precedence:
      
        times > eq# = div = eq = if > s = quot > |0| = pr > true > false
      
  
    partial status:
  
      pi(eq#) = []
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []
      pi(div) = []
      pi(quot) = []
      pi(eq) = [2]
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = []

The following rules are removed:

  p1

Then no dependency pair remains.