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, r11, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            pr#_A(x1,x2) = 1
            s_A(x1) = 23
            divides#_A(x1,x2) = max{1, x1 - 36}
            div#_A(x1,x2) = 1
            div_A(x1,x2) = x1 + 29
            zero#_A(x1) = 1
            if#_A(x1,x2,x3) = 1
            eq_A(x1,x2) = max{6, x1 - 1}
            |0|_A = 10
            plus_A(x1,x2) = max{x1 + 7, x2 + 22}
            zero_A(x1) = max{22, x1 - 30}
            false_A = 7
            divides_A(x1,x2) = 5
            times_A(x1,x2) = 8
            quot#_A(x1,x2,x3) = 1
            pr#!_A(x1,x2) = 23
            s!_A(x1) = 0
            divides#!_A(x1,x2) = 1
            div#!_A(x1,x2) = 0
            div!_A(x1,x2) = max{34, x2 + 26}
            zero#!_A(x1) = max{25, x1 + 1}
            if#!_A(x1,x2,x3) = 23
            eq!_A(x1,x2) = max{35, x1 - 2, x2 - 22}
            |0|!_A = 48
            plus!_A(x1,x2) = max{x1 - 22, x2 + 6}
            zero!_A(x1) = max{25, x1}
            false!_A = 47
            divides!_A(x1,x2) = max{x1 + 34, x2 + 33}
            times!_A(x1,x2) = max{49, x1 + 4, x2 - 22}
            quot#!_A(x1,x2,x3) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            pr#_A(x1,x2) = 1
            s_A(x1) = x1 + 97
            divides#_A(x1,x2) = 1
            div#_A(x1,x2) = 1
            div_A(x1,x2) = max{23, x1, x2 - 41}
            zero#_A(x1) = 1
            if#_A(x1,x2,x3) = 1
            eq_A(x1,x2) = max{285, x2 + 86}
            |0|_A = 94
            plus_A(x1,x2) = x1
            zero_A(x1) = max{87, x1 - 82}
            false_A = 277
            divides_A(x1,x2) = max{549, x1 + 82}
            times_A(x1,x2) = max{295, x1 - 87, x2 + 65}
            quot#_A(x1,x2,x3) = 1
            pr#!_A(x1,x2) = 0
            s!_A(x1) = x1 + 196
            divides#!_A(x1,x2) = max{x1 + 1, x2}
            div#!_A(x1,x2) = max{0, x1 - 61}
            div!_A(x1,x2) = max{x1, x2}
            zero#!_A(x1) = max{0, x1 - 101}
            if#!_A(x1,x2,x3) = 0
            eq!_A(x1,x2) = max{x1 + 66, x2 - 65}
            |0|!_A = 218
            plus!_A(x1,x2) = x2
            zero!_A(x1) = max{0, x1 - 95}
            false!_A = 164
            divides!_A(x1,x2) = max{294, x1 - 193, x2 + 65}
            times!_A(x1,x2) = max{295, x1 + 41, x2 - 97}
            quot#!_A(x1,x2,x3) = max{0, x1 - 61}
        
  
    partial status:
  
      pi(pr#) = [1]
      pi(s) = [1]
      pi(divides#) = [2]
      pi(div#) = [1]
      pi(div) = [1, 2]
      pi(zero#) = [1]
      pi(if#) = [2]
      pi(eq) = [1, 2]
      pi(|0|) = []
      pi(plus) = [1]
      pi(zero) = []
      pi(false) = []
      pi(divides) = [2]
      pi(times) = [1]
      pi(quot#) = [1]

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

  r8, r13, r17, r18, r19, r20, r21, r23, r25, r27, r30

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3) = max{0, x1 - 52, x2 - 58, x3 - 107}
            false_A = 51
            pr#_A(x1,x2) = max{0, x1 - 58, x2 - 131}
            s_A(x1) = max{51, x1 + 24}
            divides_A(x1,x2) = max{51, x1 - 81, x2 - 6}
            times_A(x1,x2) = max{0, x1 - 52}
            |0|_A = 0
            quot_A(x1,x2,x3) = max{x1 + 53, x2 - 81, x3 - 81}
            zero_A(x1) = x1 + 80
            eq_A(x1,x2) = max{51, x1 - 52}
            true_A = 51
            div_A(x1,x2) = max{x1 + 52, x2 - 80}
            pr_A(x1,x2) = max{105, x1 - 82, x2 - 108}
            if_A(x1,x2,x3) = max{105, x1 - 51, x2 - 53, x3 - 83}
            if#!_A(x1,x2,x3) = max{56, x3 - 52}
            false!_A = 31
            pr#!_A(x1,x2) = max{1, x2 - 80}
            s!_A(x1) = x1 + 33
            divides!_A(x1,x2) = max{59, x1 - 81, x2 - 106}
            times!_A(x1,x2) = max{4, x1 - 156, x2 - 79}
            |0|!_A = 3
            quot!_A(x1,x2,x3) = max{1, x1 - 82, x2 - 52, x3 - 81}
            zero!_A(x1) = x1 + 23
            eq!_A(x1,x2) = 48
            true!_A = 32
            div!_A(x1,x2) = max{2, x1 - 81, x2 - 80}
            pr!_A(x1,x2) = max{58, x1 - 107, x2 - 82}
            if!_A(x1,x2,x3) = max{32, x1 - 24, x2 + 21, x3 - 83}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 20, x3 - 127}
            false_A = 1
            pr#_A(x1,x2) = max{0, x1 - 20, x2 - 135}
            s_A(x1) = max{123, x1 + 8}
            divides_A(x1,x2) = max{1, x1 - 140, x2 - 124}
            times_A(x1,x2) = 124
            |0|_A = 115
            quot_A(x1,x2,x3) = max{x1 + 145, x2 - 124, x3 - 124}
            zero_A(x1) = max{53, x1 - 145}
            eq_A(x1,x2) = max{1, x1 - 124}
            true_A = 1
            div_A(x1,x2) = max{x1 + 35, x2 + 154}
            pr_A(x1,x2) = max{1, x1 - 135, x2 - 153}
            if_A(x1,x2,x3) = max{1, x1 - 13, x2 - 135, x3 - 144}
            if#!_A(x1,x2,x3) = max{0, x3 - 389}
            false!_A = 2
            pr#!_A(x1,x2) = max{1, x2 - 15}
            s!_A(x1) = max{0, x1 - 273}
            divides!_A(x1,x2) = max{4, x1 - 132, x2 - 440}
            times!_A(x1,x2) = max{x1 + 143, x2 + 155}
            |0|!_A = 0
            quot!_A(x1,x2,x3) = max{5, x1 - 156, x2 - 108, x3 - 1}
            zero!_A(x1) = max{5, x1 - 230}
            eq!_A(x1,x2) = max{5, x1 - 222}
            true!_A = 0
            div!_A(x1,x2) = max{x1 - 155, x2 + 154}
            pr!_A(x1,x2) = max{3, x1 - 439, x2 - 272}
            if!_A(x1,x2,x3) = max{6, x1 - 97, x2 - 440, x3 - 561}
        
  
    partial status:
  
      pi(if#) = [3]
      pi(false) = []
      pi(pr#) = [2]
      pi(s) = [1]
      pi(divides) = []
      pi(times) = []
      pi(|0|) = []
      pi(quot) = [1, 2, 3]
      pi(zero) = [1]
      pi(eq) = []
      pi(true) = []
      pi(div) = [2]
      pi(pr) = [2]
      pi(if) = [1, 2, 3]

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: 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, r5, r6, r7, r8, r11, r13, r27

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{1, x1 - 12}
            div_A(x1,x2) = max{11, x1 + 5, x2 + 10}
            times_A(x1,x2) = max{13, x1 - 52}
            zero_A(x1) = x1 + 42
            quot#_A(x1,x2,x3) = max{1, x1 - 12}
            |0|_A = 0
            s_A(x1) = 32
            p_A(x1) = max{42, x1 - 32}
            plus_A(x1,x2) = max{x1 + 52, x2 + 10}
            quot_A(x1,x2,x3) = max{12, x2 - 43, x3 - 43}
            div#!_A(x1,x2) = max{52, x1 + 1}
            div!_A(x1,x2) = max{68, x1 - 2, x2 + 54}
            times!_A(x1,x2) = max{7, x1 - 43, x2 - 21}
            zero!_A(x1) = max{53, x1 - 36}
            quot#!_A(x1,x2,x3) = max{52, x1 + 1}
            |0|!_A = 4
            s!_A(x1) = 5
            p!_A(x1) = 10
            plus!_A(x1,x2) = max{x1 - 47, x2 + 10}
            quot!_A(x1,x2,x3) = max{6, x2 - 33}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{1, x1 - 12}
            div_A(x1,x2) = max{x1 + 11, x2 - 20}
            times_A(x1,x2) = max{12, x1 - 20, x2 - 17}
            zero_A(x1) = x1 + 3
            quot#_A(x1,x2,x3) = max{1, x1 - 12}
            |0|_A = 0
            s_A(x1) = 0
            p_A(x1) = max{3, x1 - 17}
            plus_A(x1,x2) = x1
            quot_A(x1,x2,x3) = max{12, x1 - 20, x2 - 20, x3 - 20}
            div#!_A(x1,x2) = 0
            div!_A(x1,x2) = max{17, x1 - 10, x2 - 20}
            times!_A(x1,x2) = max{18, x1 - 20}
            zero!_A(x1) = max{0, x1 - 20}
            quot#!_A(x1,x2,x3) = 0
            |0|!_A = 15
            s!_A(x1) = 2
            p!_A(x1) = max{2, x1 - 20}
            plus!_A(x1,x2) = max{2, x1 - 20}
            quot!_A(x1,x2,x3) = max{16, x1 + 11, x2 - 20, x3 - 20}
        
  
    partial status:
  
      pi(div#) = [1]
      pi(div) = [1, 2]
      pi(times) = []
      pi(zero) = [1]
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = []
      pi(p) = []
      pi(plus) = [1]
      pi(quot) = [1, 3]

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

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{3, x1 - 1, x2 - 4}
            quot#_A(x1,x2,x3) = max{3, x1 - 1, x2 - 31, x3 - 4}
            |0|_A = 30
            s_A(x1) = 6
            p_A(x1) = x1 + 31
            div#!_A(x1,x2) = x2 + 24
            quot#!_A(x1,x2,x3) = max{1, x2, x3 - 5}
            |0|!_A = 24
            s!_A(x1) = 22
            p!_A(x1) = max{21, x1 - 4}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{21, x1 - 1, x2 + 6}
            quot#_A(x1,x2,x3) = max{21, x1 - 1, x2 - 21, x3 + 6}
            |0|_A = 20
            s_A(x1) = 14
            p_A(x1) = max{60, x1 + 48}
            div#!_A(x1,x2) = max{151, x2 + 132}
            quot#!_A(x1,x2,x3) = max{145, x2 + 131}
            |0|!_A = 152
            s!_A(x1) = max{150, x1 + 58}
            p!_A(x1) = max{153, x1 + 6}
        
  
    partial status:
  
      pi(div#) = [2]
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(p) = []

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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = max{0, x1 - 27}
            s_A(x1) = max{65, x1 + 26}
            div_A(x1,x2) = max{721, x1 - 1165, x2 - 92}
            |0|_A = 363
            times#!_A(x1,x2) = max{1, x1 - 66, x2 - 1}
            s!_A(x1) = max{67, x1 + 52}
            div!_A(x1,x2) = max{54, x2 + 6}
            |0|!_A = 53
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = max{0, x1 - 20}
            s_A(x1) = max{48, x1 + 19}
            div_A(x1,x2) = max{668, x1 - 903, x2 - 68}
            |0|_A = 213
            times#!_A(x1,x2) = max{10, x1 - 49, x2 - 1}
            s!_A(x1) = max{9, x1 + 5}
            div!_A(x1,x2) = max{2, x1 - 2863, x2 - 14}
            |0|!_A = 4
        
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = [1]
      pi(div) = [1, 2]
      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

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 33, x2 - 33}
            s_A(x1) = max{32, x1 + 16}
            p_A(x1) = max{17, x1 - 16}
            |0|_A = 6
            plus#!_A(x1,x2) = max{1, x1 - 30}
            s!_A(x1) = max{12, x1 + 8}
            p!_A(x1) = max{0, x1 - 7}
            |0|!_A = 10
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 29, x2 - 29}
            s_A(x1) = x1 + 21
            p_A(x1) = max{28, x1 - 10}
            |0|_A = 9
            plus#!_A(x1,x2) = max{6, x1 - 29, x2 - 22}
            s!_A(x1) = max{5, x1 - 342}
            p!_A(x1) = max{0, x1 - 46}
            |0|!_A = 2
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(p) = [1]
      pi(|0|) = []

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

  r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 17, x2 - 47}
            s_A(x1) = max{46, x1 + 16}
            p_A(x1) = max{31, x1 - 9}
            plus#!_A(x1,x2) = max{4, x2 - 40}
            s!_A(x1) = max{5, x1 - 434}
            p!_A(x1) = max{3, x1 - 431}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 25, x2 - 24}
            s_A(x1) = x1 + 23
            p_A(x1) = max{0, x1 - 9}
            plus#!_A(x1,x2) = max{7, x2 - 15}
            s!_A(x1) = max{0, x1 - 5}
            p!_A(x1) = max{126, x1 - 17}
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(p) = [1]

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = max{0, x1 - 14, x2 - 14}
            s_A(x1) = max{41, x1 + 13}
            eq#!_A(x1,x2) = max{86, x1 - 83, x2 - 42}
            s!_A(x1) = max{85, x1 - 83}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = max{0, x1 - 15, x2 - 15}
            s_A(x1) = max{44, x1 + 14}
            eq#!_A(x1,x2) = max{23, x1 - 226, x2 - 45}
            s!_A(x1) = max{22, x1 - 227}
        
  
    partial status:
  
      pi(eq#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.