YES
We show the termination of the TRS R:

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

The set of usable rules consists of

  r3, r8, r10, r16, r22, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if#_A(x1,x2,x3) = x3
          false_A() = 3
          pr#_A(x1,x2) = x2
          s_A(x1) = x1
          divides_A(x1,x2) = 1
          plus_A(x1,x2) = x2 + 2
          |0|_A() = 2
          times_A(x1,x2) = x2 + 4
          div_A(x1,x2) = 3
          eq_A(x1,x2) = x2
          true_A() = 1
          pr_A(x1,x2) = x2 + 5
          if_A(x1,x2,x3) = 4
  
    base order 2:
  
      precedence:
      
        |0| > s = plus = pr > if# = div = true = if > pr# = times = eq > false = divides
      
  
    partial status:
  
      pi(if#) = [3]
      pi(false) = []
      pi(pr#) = [2]
      pi(s) = [1]
      pi(divides) = []
      pi(plus) = []
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      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(s(x)) -> x
r2: plus(x,|0|()) -> x
r3: plus(|0|(),y) -> y
r4: plus(s(x),y) -> s(plus(x,y))
r5: plus(s(x),y) -> s(plus(p(s(x)),y))
r6: plus(x,s(y)) -> s(plus(x,p(s(y))))
r7: times(|0|(),y) -> |0|()
r8: times(s(|0|()),y) -> y
r9: times(s(x),y) -> plus(y,times(x,y))
r10: div(|0|(),y) -> |0|()
r11: div(x,y) -> quot(x,y,y)
r12: quot(|0|(),s(y),z) -> |0|()
r13: quot(s(x),s(y),z) -> quot(x,y,z)
r14: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r15: div(div(x,y),z) -> div(x,times(y,z))
r16: eq(|0|(),|0|()) -> true()
r17: eq(s(x),|0|()) -> false()
r18: eq(|0|(),s(y)) -> false()
r19: eq(s(x),s(y)) -> eq(x,y)
r20: divides(y,x) -> eq(x,times(div(x,y),y))
r21: prime(s(s(x))) -> pr(s(s(x)),s(x))
r22: pr(x,s(|0|())) -> true()
r23: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r24: if(true(),x,y) -> false()
r25: 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(s(x)) -> x
r2: plus(x,|0|()) -> x
r3: plus(|0|(),y) -> y
r4: plus(s(x),y) -> s(plus(x,y))
r5: plus(s(x),y) -> s(plus(p(s(x)),y))
r6: plus(x,s(y)) -> s(plus(x,p(s(y))))
r7: times(|0|(),y) -> |0|()
r8: times(s(|0|()),y) -> y
r9: times(s(x),y) -> plus(y,times(x,y))
r10: div(|0|(),y) -> |0|()
r11: div(x,y) -> quot(x,y,y)
r12: quot(|0|(),s(y),z) -> |0|()
r13: quot(s(x),s(y),z) -> quot(x,y,z)
r14: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r15: div(div(x,y),z) -> div(x,times(y,z))
r16: eq(|0|(),|0|()) -> true()
r17: eq(s(x),|0|()) -> false()
r18: eq(|0|(),s(y)) -> false()
r19: eq(s(x),s(y)) -> eq(x,y)
r20: divides(y,x) -> eq(x,times(div(x,y),y))
r21: prime(s(s(x))) -> pr(s(s(x)),s(x))
r22: pr(x,s(|0|())) -> true()
r23: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r24: if(true(),x,y) -> false()
r25: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r2, r10, r12, r16, r17, r18, r22, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          div#_A(x1,x2) = x1 + 2
          div_A(x1,x2) = x1 + x2 + 3
          times_A(x1,x2) = 0
          quot#_A(x1,x2,x3) = x1 + 2
          |0|_A() = 1
          s_A(x1) = x1 + 3
          plus_A(x1,x2) = x1 + 4
          quot_A(x1,x2,x3) = 2
          eq_A(x1,x2) = x2 + 5
          true_A() = 5
          false_A() = 4
          pr_A(x1,x2) = 7
          if_A(x1,x2,x3) = 8
  
    base order 2:
  
      precedence:
      
        times > div# = quot# = eq > div = true = if > quot = false > s = plus = pr > |0|
      
  
    partial status:
  
      pi(div#) = [1]
      pi(div) = []
      pi(times) = []
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = []
      pi(plus) = [1]
      pi(quot) = []
      pi(eq) = [2]
      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: 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))

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  r2, r3, r10, r12, r16, r17, r18, r19, r22, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          div#_A(x1,x2) = x1
          div_A(x1,x2) = x1 + 4
          times_A(x1,x2) = 0
          quot#_A(x1,x2,x3) = x1
          |0|_A() = 3
          s_A(x1) = x1 + 6
          plus_A(x1,x2) = x1 + x2 + 7
          quot_A(x1,x2,x3) = x1 + x3 + 7
          eq_A(x1,x2) = x1
          true_A() = 2
          false_A() = 1
          pr_A(x1,x2) = x2
          if_A(x1,x2,x3) = 2
  
    base order 2:
  
      precedence:
      
        quot > |0| = eq > div# = quot# = plus = true = if > div = times = s = false > pr
      
  
    partial status:
  
      pi(div#) = []
      pi(div) = []
      pi(times) = []
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(plus) = [1]
      pi(quot) = [3]
      pi(eq) = [1]
      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))

and R consists of:

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

The set of usable rules consists of

  r3, r4, r5, r6, r7, r10, r12, r14, r16, r17, r18, r22, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          div#_A(x1,x2) = x2 + 1
          quot#_A(x1,x2,x3) = x2
          |0|_A() = 5
          s_A(x1) = 3
          plus_A(x1,x2) = x2 + 4
          p_A(x1) = 1
          times_A(x1,x2) = x1 + 13
          div_A(x1,x2) = x1 + 6
          quot_A(x1,x2,x3) = x1 + 7
          eq_A(x1,x2) = x1 + x2 + 4
          true_A() = 2
          false_A() = 1
          pr_A(x1,x2) = 4
          if_A(x1,x2,x3) = x2 + 5
  
    base order 2:
  
      precedence:
      
        div# = times = quot = if > quot# = div = eq = pr > s > plus = true > p = false > |0|
      
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(plus) = [2]
      pi(p) = []
      pi(times) = []
      pi(div) = [1]
      pi(quot) = [1]
      pi(eq) = [1, 2]
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = [2]

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)

and R consists of:

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

The set of usable rules consists of

  r2, r3, r8, r10, r12, r16, r17, r18, r21, r22, r23, r24, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          times#_A(x1,x2) = x1
          s_A(x1) = x1 + 5
          plus_A(x1,x2) = x1 + x2 + 6
          |0|_A() = 1
          times_A(x1,x2) = x1 + x2 + 6
          div_A(x1,x2) = x1 + x2 + 7
          quot_A(x1,x2,x3) = x1 + x3 + 13
          eq_A(x1,x2) = 4
          true_A() = 3
          false_A() = 2
          prime_A(x1) = x1 + 1
          pr_A(x1,x2) = 4
          if_A(x1,x2,x3) = 4
          divides_A(x1,x2) = x1
  
    base order 2:
  
      precedence:
      
        plus > times = quot > div = prime > s > times# = eq > true = divides > |0| > false > pr = if
      
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = [1]
      pi(plus) = [1]
      pi(|0|) = []
      pi(times) = []
      pi(div) = [2]
      pi(quot) = [3]
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(prime) = [1]
      pi(pr) = []
      pi(if) = []
      pi(divides) = []

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

The set of usable rules consists of

  r1, r2, r3, r7, r10, r12, r16, r17, r18, r19, r22, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1
          s_A(x1) = x1 + 6
          p_A(x1) = x1
          plus_A(x1,x2) = x1 + x2 + 7
          |0|_A() = 3
          times_A(x1,x2) = 5
          div_A(x1,x2) = 4
          quot_A(x1,x2,x3) = x1 + x3 + 7
          eq_A(x1,x2) = x1 + x2 + 7
          true_A() = 2
          false_A() = 1
          pr_A(x1,x2) = x1 + x2 + 7
          if_A(x1,x2,x3) = x3 + 2
  
    base order 2:
  
      precedence:
      
        pr > plus = quot > s > plus# = times > p = div > |0| > true > eq > false = if
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(p) = []
      pi(plus) = [2]
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(quot) = [3]
      pi(eq) = [1, 2]
      pi(true) = []
      pi(false) = []
      pi(pr) = []
      pi(if) = [3]

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

The set of usable rules consists of

  r2, r3, r7, r10, r12, r16, r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1 + 1
          s_A(x1) = x1
          p_A(x1) = x1 + 2
          plus_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 3
          times_A(x1,x2) = x1 + 5
          div_A(x1,x2) = x1 + 2
          quot_A(x1,x2,x3) = x1 + x2 + x3 + 3
          eq_A(x1,x2) = x1 + x2 + 1
          true_A() = 2
          false_A() = 1
  
    base order 2:
  
      precedence:
      
        quot > times = div > |0| = eq > plus# = s = p = plus > true > false
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(p) = []
      pi(plus) = [1]
      pi(|0|) = []
      pi(times) = []
      pi(div) = [1]
      pi(quot) = [3]
      pi(eq) = [2]
      pi(true) = []
      pi(false) = []

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

The set of usable rules consists of

  r1, r2, r3, r7, r10, r12, r16, r17, r18, r22, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x2
          s_A(x1) = x1 + 5
          p_A(x1) = x1
          plus_A(x1,x2) = x1 + x2 + 4
          |0|_A() = 4
          times_A(x1,x2) = x1 + 25
          div_A(x1,x2) = x1 + 12
          quot_A(x1,x2,x3) = x1 + x3 + 6
          eq_A(x1,x2) = x1 + x2 + 6
          true_A() = 1
          false_A() = 3
          pr_A(x1,x2) = 2
          if_A(x1,x2,x3) = x3 + 4
  
    base order 2:
  
      precedence:
      
        plus = false > times = eq > s = div = quot = if > plus# = |0| = pr > p = true
      
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = [1]
      pi(p) = []
      pi(plus) = [1]
      pi(|0|) = []
      pi(times) = []
      pi(div) = [1]
      pi(quot) = [3]
      pi(eq) = [1]
      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(s(x)) -> x
r2: plus(x,|0|()) -> x
r3: plus(|0|(),y) -> y
r4: plus(s(x),y) -> s(plus(x,y))
r5: plus(s(x),y) -> s(plus(p(s(x)),y))
r6: plus(x,s(y)) -> s(plus(x,p(s(y))))
r7: times(|0|(),y) -> |0|()
r8: times(s(|0|()),y) -> y
r9: times(s(x),y) -> plus(y,times(x,y))
r10: div(|0|(),y) -> |0|()
r11: div(x,y) -> quot(x,y,y)
r12: quot(|0|(),s(y),z) -> |0|()
r13: quot(s(x),s(y),z) -> quot(x,y,z)
r14: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r15: div(div(x,y),z) -> div(x,times(y,z))
r16: eq(|0|(),|0|()) -> true()
r17: eq(s(x),|0|()) -> false()
r18: eq(|0|(),s(y)) -> false()
r19: eq(s(x),s(y)) -> eq(x,y)
r20: divides(y,x) -> eq(x,times(div(x,y),y))
r21: prime(s(s(x))) -> pr(s(s(x)),s(x))
r22: pr(x,s(|0|())) -> true()
r23: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r24: if(true(),x,y) -> false()
r25: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r2, r7, r10, r12, r16, r17, r18, r21, r22, r23, r24, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          eq#_A(x1,x2) = x1 + x2 + 1
          s_A(x1) = x1
          plus_A(x1,x2) = x1 + 2
          |0|_A() = 2
          times_A(x1,x2) = x1 + 7
          div_A(x1,x2) = x1 + 3
          quot_A(x1,x2,x3) = x1 + x3 + 4
          eq_A(x1,x2) = x1 + x2 + 2
          true_A() = 1
          false_A() = 3
          prime_A(x1) = x1 + 5
          pr_A(x1,x2) = 4
          if_A(x1,x2,x3) = 4
          divides_A(x1,x2) = x1
  
    base order 2:
  
      precedence:
      
        times = quot > div = false > eq# = plus = |0| = eq = prime > s = true = pr = if > divides
      
  
    partial status:
  
      pi(eq#) = [1]
      pi(s) = [1]
      pi(plus) = []
      pi(|0|) = []
      pi(times) = [1]
      pi(div) = [1]
      pi(quot) = [3]
      pi(eq) = []
      pi(true) = []
      pi(false) = []
      pi(prime) = [1]
      pi(pr) = []
      pi(if) = []
      pi(divides) = []

The following rules are removed:

  p1

Then no dependency pair remains.