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, r11, r13, r25, r27, r28

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        pr#_A(x1,x2) = max{x1 - 54, x2 + 140}
        s_A(x1) = 0
        divides#_A(x1,x2) = max{140, x1 + 14}
        div#_A(x1,x2) = 140
        div_A(x1,x2) = x1 + 6
        zero#_A(x1) = 140
        if#_A(x1,x2,x3) = max{x2 - 54, x3 + 140}
        eq_A(x1,x2) = max{x1 + 9, x2 - 10}
        |0|_A = 19
        plus_A(x1,x2) = max{120, x1 + 53, x2 + 34}
        zero_A(x1) = max{86, x1 + 13}
        false_A = 87
        divides_A(x1,x2) = max{141, x1 - 14, x2 + 16}
        times_A(x1,x2) = max{11, x1 + 7, x2 - 5}
        quot#_A(x1,x2,x3) = 140
        quot_A(x1,x2,x3) = x1 + 8
        if_A(x1,x2,x3) = max{86, x1 - 2, x2 - 140, x3 - 113}
        true_A = 90
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          pr#_A(x1,x2) = max{71, x2 - 207}
          s_A(x1) = max{172, x1 + 34}
          divides#_A(x1,x2) = 71
          div#_A(x1,x2) = 71
          div_A(x1,x2) = max{x1 + 168, x2 + 45}
          zero#_A(x1) = 71
          if#_A(x1,x2,x3) = max{71, x3 - 173}
          eq_A(x1,x2) = max{171, x1 - 1, x2 - 245}
          |0|_A = 71
          plus_A(x1,x2) = max{73, x1 + 1, x2}
          zero_A(x1) = x1 + 139
          false_A = 172
          divides_A(x1,x2) = max{41, x1 + 1, x2 - 140}
          times_A(x1,x2) = max{0, x1 - 116}
          quot#_A(x1,x2,x3) = 71
          quot_A(x1,x2,x3) = max{71, x1 - 69, x2 - 173, x3 - 1}
          if_A(x1,x2,x3) = max{174, x1 - 174, x3 - 173}
          true_A = 214
          pr#!_A(x1,x2) = max{126, x1 - 45}
          s!_A(x1) = 123
          divides#!_A(x1,x2) = max{125, x2 - 46}
          div#!_A(x1,x2) = max{124, x1 - 47}
          div!_A(x1,x2) = max{174, x1 - 271}
          zero#!_A(x1) = max{27, x1 - 3}
          if#!_A(x1,x2,x3) = max{126, x2 - 45}
          eq!_A(x1,x2) = max{172, x1 - 43, x2 - 51}
          |0|!_A = 30
          plus!_A(x1,x2) = 0
          zero!_A(x1) = max{122, x1 - 1}
          false!_A = 0
          divides!_A(x1,x2) = max{173, x1 - 84, x2 - 44}
          times!_A(x1,x2) = max{0, x1 - 172, x2 - 172}
          quot#!_A(x1,x2,x3) = max{124, x1 - 47}
          quot!_A(x1,x2,x3) = max{174, x1 - 139, x2 - 3}
          if!_A(x1,x2,x3) = max{172, x1 - 213, x2 - 71, x3 - 212}
          true!_A = 29
  
    partial status:
  
      pi(pr#) = []
      pi(s) = []
      pi(divides#) = []
      pi(div#) = []
      pi(div) = [1]
      pi(zero#) = []
      pi(if#) = []
      pi(eq) = []
      pi(|0|) = []
      pi(plus) = [1]
      pi(zero) = [1]
      pi(false) = []
      pi(divides) = [2]
      pi(times) = []
      pi(quot#) = []
      pi(quot) = [1]
      pi(if) = []
      pi(true) = []

The following rules are removed:

  p1, p2, p3, p4, p10



-- SCC decomposition.

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))
p3: div#(div(x,y),z) -> div#(x,times(zero(y),z))
p4: div#(x,y) -> quot#(x,y,y)
p5: 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(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, p4, p5}


-- 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

  r11, r17, r18, r19, r20, r23, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if#_A(x1,x2,x3) = x3
        false_A = 0
        pr#_A(x1,x2) = x2
        s_A(x1) = x1
        divides_A(x1,x2) = 9
        div_A(x1,x2) = max{14, x1 - 14}
        |0|_A = 0
        eq_A(x1,x2) = max{8, x1 + 6, x2 + 3}
        true_A = 7
        pr_A(x1,x2) = max{x1 - 8, x2 + 8}
        if_A(x1,x2,x3) = max{x1 - 10, x2 - 7, x3 + 9}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if#_A(x1,x2,x3) = max{29, x2, x3 - 58}
          false_A = 24
          pr#_A(x1,x2) = max{29, x1, x2 - 58}
          s_A(x1) = max{57, x1 + 29}
          divides_A(x1,x2) = max{110, x1 - 31, x2 + 55}
          div_A(x1,x2) = 111
          |0|_A = 111
          eq_A(x1,x2) = 24
          true_A = 24
          pr_A(x1,x2) = max{24, x1 - 95, x2 - 141}
          if_A(x1,x2,x3) = max{24, x1 - 110, x2 - 95, x3 - 111}
          if#!_A(x1,x2,x3) = 134
          false!_A = 59
          pr#!_A(x1,x2) = 134
          s!_A(x1) = 135
          divides!_A(x1,x2) = max{x1 + 79, x2 + 166}
          div!_A(x1,x2) = 0
          |0|!_A = 60
          eq!_A(x1,x2) = max{62, x1 + 1, x2 + 34}
          true!_A = 166
          pr!_A(x1,x2) = max{x1 + 165, x2 - 141}
          if!_A(x1,x2,x3) = max{x1 + 111, x2 - 95}
  
    partial status:
  
      pi(if#) = [3]
      pi(false) = []
      pi(pr#) = [2]
      pi(s) = [1]
      pi(divides) = []
      pi(div) = []
      pi(|0|) = []
      pi(eq) = [1, 2]
      pi(true) = []
      pi(pr) = [2]
      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(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: div#(div(x,y),z) -> div#(x,times(zero(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(|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, r27, r30

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = max{10, x1 + 1}
        div_A(x1,x2) = max{13, x1 + 2, x2 - 10}
        times_A(x1,x2) = max{8, x1 - 28, x2 - 4}
        zero_A(x1) = max{27, x1 + 21}
        quot#_A(x1,x2,x3) = max{10, x1 + 1}
        |0|_A = 6
        s_A(x1) = 10
        p_A(x1) = 8
        quot_A(x1,x2,x3) = max{13, x2 - 5, x3 - 10}
        eq_A(x1,x2) = max{12, x1 - 10, x2 - 16}
        true_A = 11
        false_A = 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{1, x1 - 49}
          div_A(x1,x2) = max{x1 + 48, x2 - 82}
          times_A(x1,x2) = 49
          zero_A(x1) = max{6, x1 - 20}
          quot#_A(x1,x2,x3) = max{1, x1 - 49}
          |0|_A = 26
          s_A(x1) = max{0, x1 - 23}
          p_A(x1) = max{27, x1 + 12}
          quot_A(x1,x2,x3) = max{25, x1 + 24, x2 - 132, x3 - 106}
          eq_A(x1,x2) = max{x1 - 27, x2 + 11}
          true_A = 9
          false_A = 9
          div#!_A(x1,x2) = x1 + 84
          div!_A(x1,x2) = max{0, x1 - 49, x2 - 51}
          times!_A(x1,x2) = max{x1 + 22, x2 - 1}
          zero!_A(x1) = 0
          quot#!_A(x1,x2,x3) = x1 + 84
          |0|!_A = 0
          s!_A(x1) = 0
          p!_A(x1) = max{1, x1 - 27}
          quot!_A(x1,x2,x3) = max{1, x1 - 52, x2 - 78, x3 - 1}
          eq!_A(x1,x2) = max{72, x1 - 27, x2 - 119}
          true!_A = 1
          false!_A = 70
  
    partial status:
  
      pi(div#) = [1]
      pi(div) = []
      pi(times) = []
      pi(zero) = []
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = []
      pi(p) = []
      pi(quot) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []

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(|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: 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(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

  r23

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = x2 + 12
        quot#_A(x1,x2,x3) = max{11, x2 + 1}
        |0|_A = 36
        s_A(x1) = 23
        pr_A(x1,x2) = max{22, x1 - 20, x2 - 33}
        true_A = 21
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{0, x1 - 2, x2 - 5}
          quot#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 210, x3 - 5}
          |0|_A = 208
          s_A(x1) = max{4, x1 - 61}
          pr_A(x1,x2) = max{279, x1 - 229, x2 - 136}
          true_A = 17
          div#!_A(x1,x2) = 0
          quot#!_A(x1,x2,x3) = x2 + 63
          |0|!_A = 4
          s!_A(x1) = max{0, x1 - 270}
          pr!_A(x1,x2) = max{0, x1 - 491, x2 - 148}
          true!_A = 3
  
    partial status:
  
      pi(div#) = [2]
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(pr) = []
      pi(true) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- 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

  r11

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = x2 + 2
        s_A(x1) = max{0, x1 - 50}
        div_A(x1,x2) = max{4, x1 - 44, x2 - 8}
        |0|_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times#_A(x1,x2) = max{0, x1 - 6}
          s_A(x1) = max{17, x1 + 5}
          div_A(x1,x2) = max{191, x1 - 291, x2 - 23}
          |0|_A = 95
          times#!_A(x1,x2) = max{6, x1 - 6}
          s!_A(x1) = max{5, x1 - 1668}
          div!_A(x1,x2) = max{7, x1 - 710, x2 - 18}
          |0|!_A = 2
  
    partial status:
  
      pi(times#) = []
      pi(s) = []
      pi(div) = []
      pi(|0|) = []

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

  r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = 1
        s_A(x1) = max{9, x1 - 70}
        p_A(x1) = max{163, x1 + 73}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 47, x2 - 47}
          s_A(x1) = max{46, x1 + 34}
          p_A(x1) = max{20, x1 - 30}
          plus#!_A(x1,x2) = max{2, x1 - 43}
          s!_A(x1) = max{3, x1 - 1179}
          p!_A(x1) = max{3, x1 - 1213}
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      pi(p) = []

The following rules are removed:

  p1, p3



-- 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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1 + 1
        s_A(x1) = max{7, x1 - 76}
        p_A(x1) = max{184, x1 + 179}
        |0|_A = 3
        times_A(x1,x2) = max{x1 - 82, x2 + 12}
        div_A(x1,x2) = max{81, x1 + 41, x2 - 53}
        quot_A(x1,x2,x3) = max{4, x1 - 259, x2 - 8, x3 - 1}
        zero_A(x1) = max{258, x1 - 11}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 5, x2 - 5}
          s_A(x1) = x1 + 4
          p_A(x1) = max{1, x1 - 3}
          |0|_A = 0
          times_A(x1,x2) = max{14, x1 - 6, x2 - 2}
          div_A(x1,x2) = max{11, x1 - 20, x2 - 20}
          quot_A(x1,x2,x3) = max{3, x1 - 19, x2 - 13, x3 - 17}
          zero_A(x1) = max{18, x1 - 65}
          plus#!_A(x1,x2) = max{1, x2 - 2}
          s!_A(x1) = max{7, x1 - 109}
          p!_A(x1) = max{5, x1 - 38}
          |0|!_A = 4
          times!_A(x1,x2) = max{10, x2 - 42}
          div!_A(x1,x2) = max{9, x1 - 132, x2 - 134}
          quot!_A(x1,x2,x3) = max{8, x1 - 129, x2 - 5, x3 - 101}
          zero!_A(x1) = max{0, x1 - 16}
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(p) = []
      pi(|0|) = []
      pi(times) = [2]
      pi(div) = [1]
      pi(quot) = []
      pi(zero) = []

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

  r11

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        eq#_A(x1,x2) = 0
        s_A(x1) = max{15, x1 - 105}
        div_A(x1,x2) = max{13, x1 - 54, x2 - 51}
        |0|_A = 12
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          eq#_A(x1,x2) = max{0, x1 - 4, x2 - 4}
          s_A(x1) = max{15, x1 + 3}
          div_A(x1,x2) = max{181, x1 - 290, x2 - 19}
          |0|_A = 91
          eq#!_A(x1,x2) = max{10, x1 - 4, x2 - 16}
          s!_A(x1) = max{9, x1 - 1179}
          div!_A(x1,x2) = max{11, x1 - 997, x2 - 299}
          |0|!_A = 3
  
    partial status:
  
      pi(eq#) = []
      pi(s) = []
      pi(div) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.