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

  r10, r16, r17, r18, r19, r20

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if#_A(x1,x2,x3) = x3
              false_A() = 1
              pr#_A(x1,x2) = x2
              s_A(x1) = x1 + 3
              divides_A(x1,x2) = x1 + 1
              div_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 4
              eq_A(x1,x2) = 1
              true_A() = 0
              times_A(x1,x2) = 2
              if#!_A(x1,x2,x3) = 10
              false!_A() = 1
              pr#!_A(x1,x2) = 9
              s!_A(x1) = 10
              divides!_A(x1,x2) = 8
              div!_A(x1,x2) = 12
              |0|!_A() = 11
              eq!_A(x1,x2) = 3
              true!_A() = 2
              times!_A(x1,x2) = x1 + 11
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3) = max{1, x1 - 12, x2 - 12, x3 - 3}
            false_A = 7
            pr#_A(x1,x2) = max{1, x1 - 12, x2 - 4}
            s_A(x1) = 2
            divides_A(x1,x2) = x1 + 8
            div_A(x1,x2) = max{7, x1 + 5, x2 - 12}
            |0|_A = 8
            eq_A(x1,x2) = 7
            true_A = 7
            times_A(x1,x2) = max{6, x2 - 14}
            if#!_A(x1,x2,x3) = max{0, x1 - 12, x3 - 3}
            false!_A = 0
            pr#!_A(x1,x2) = max{0, x2 - 12}
            s!_A(x1) = 0
            divides!_A(x1,x2) = 0
            div!_A(x1,x2) = max{x1 + 33, x2 + 14}
            |0|!_A = 17
            eq!_A(x1,x2) = 0
            true!_A = 1
            times!_A(x1,x2) = max{18, x2 - 4}
        
  
    partial status:
  
      pi(if#) = [3]
      pi(false) = []
      pi(pr#) = [2]
      pi(s) = [1]
      pi(divides) = []
      pi(div) = [2]
      pi(|0|) = []
      pi(eq) = []
      pi(true) = []
      pi(times) = [1]

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(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

  r3, r8, r12

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              div#_A(x1,x2) = x1
              div_A(x1,x2) = x1 + x2 + 1
              times_A(x1,x2) = x2 + 2
              quot#_A(x1,x2,x3) = x1
              |0|_A() = 3
              s_A(x1) = x1 + 2
              plus_A(x1,x2) = x2 + 1
              quot_A(x1,x2,x3) = x1 + 2
              div#!_A(x1,x2) = x1 + 5
              div!_A(x1,x2) = 4
              times!_A(x1,x2) = 7
              quot#!_A(x1,x2,x3) = x1 + 5
              |0|!_A() = 2
              s!_A(x1) = 2
              plus!_A(x1,x2) = 1
              quot!_A(x1,x2,x3) = x1 + x3 + 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = 1
            div_A(x1,x2) = max{18, x1 + 8, x2 + 8}
            times_A(x1,x2) = max{x1 - 16, x2}
            quot#_A(x1,x2,x3) = 1
            |0|_A = 0
            s_A(x1) = max{0, x1 - 7}
            plus_A(x1,x2) = max{8, x2}
            quot_A(x1,x2,x3) = max{0, x1 - 1, x2 - 1, x3 - 1}
            div#!_A(x1,x2) = x1 + 6
            div!_A(x1,x2) = 3
            times!_A(x1,x2) = max{1, x1 - 19, x2 - 10}
            quot#!_A(x1,x2,x3) = x1 + 6
            |0|!_A = 2
            s!_A(x1) = 4
            plus!_A(x1,x2) = max{22, x1 + 21, x2 - 9}
            quot!_A(x1,x2,x3) = max{x1 + 15, x2 + 5, x3 + 15}
        
  
    partial status:
  
      pi(div#) = []
      pi(div) = []
      pi(times) = [2]
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []
      pi(plus) = [1, 2]
      pi(quot) = []

The following rules are removed:

  p1, p4



-- 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

  r7

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              div#_A(x1,x2) = 0
              quot#_A(x1,x2,x3) = 0
              |0|_A() = 7
              s_A(x1) = 3
              times_A(x1,x2) = x1 + 1
              div#!_A(x1,x2) = x1 + x2 + 1
              quot#!_A(x1,x2,x3) = x1 + x2
              |0|!_A() = 5
              s!_A(x1) = x1 + 6
              times!_A(x1,x2) = x2 + 16
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{1, x1 - 2, x2 - 49}
            quot#_A(x1,x2,x3) = max{1, x1 - 2, x2 - 49}
            |0|_A = 30
            s_A(x1) = 48
            times_A(x1,x2) = max{55, x1 + 6, x2 + 53}
            div#!_A(x1,x2) = 29
            quot#!_A(x1,x2,x3) = 0
            |0|!_A = 28
            s!_A(x1) = 0
            times!_A(x1,x2) = max{26, x1 + 6, x2 - 53}
        
  
    partial status:
  
      pi(div#) = [1]
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = []
      pi(times) = [1, 2]

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(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

  r7

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              times#_A(x1,x2) = x1
              s_A(x1) = x1 + 3
              times_A(x1,x2) = x1 + x2 + 5
              |0|_A() = 6
              times#!_A(x1,x2) = x1
              s!_A(x1) = 6
              times!_A(x1,x2) = 7
              |0|!_A() = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = 0
            s_A(x1) = max{8, x1 - 18}
            times_A(x1,x2) = max{10, x1 + 4, x2 - 124}
            |0|_A = 11
            times#!_A(x1,x2) = max{22, x1 - 9, x2 - 1}
            s!_A(x1) = x1 + 21
            times!_A(x1,x2) = max{81, x2 + 4}
            |0|!_A = 1
        
  
    partial status:
  
      pi(times#) = []
      pi(s) = [1]
      pi(times) = []
      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(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

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x2
              s_A(x1) = x1 + 2
              p_A(x1) = x1
              plus#!_A(x1,x2) = x1 + 1
              s!_A(x1) = x1 + 4
              p!_A(x1) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = x1 + 53
            p_A(x1) = max{1, x1 - 50}
            plus#!_A(x1,x2) = x2 + 56
            s!_A(x1) = 46
            p!_A(x1) = 55
        
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = [1]
      pi(p) = [1]

The following rules are removed:

  p1, p2



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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, r7, r10, r12

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x1
              s_A(x1) = x1 + 4
              p_A(x1) = x1
              times_A(x1,x2) = x1 + 1
              |0|_A() = 1
              div_A(x1,x2) = x1 + 6
              quot_A(x1,x2,x3) = x1 + 7
              plus#!_A(x1,x2) = 7
              s!_A(x1) = 7
              p!_A(x1) = x1 + 2
              times!_A(x1,x2) = x1
              |0|!_A() = 0
              div!_A(x1,x2) = x2 + 4
              quot!_A(x1,x2,x3) = 7
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{11, x1 - 117, x2 - 2}
            s_A(x1) = max{157, x1 + 42}
            p_A(x1) = max{129, x1 - 41}
            times_A(x1,x2) = max{155, x1 - 162}
            |0|_A = 1
            div_A(x1,x2) = max{242, x1 + 39, x2 - 40}
            quot_A(x1,x2,x3) = max{243, x1 - 2, x2 - 158, x3 - 158}
            plus#!_A(x1,x2) = max{12, x1 - 2}
            s!_A(x1) = x1 + 40
            p!_A(x1) = max{156, x1 - 158}
            times!_A(x1,x2) = max{2, x1 - 244}
            |0|!_A = 1
            div!_A(x1,x2) = max{2, x2 - 118}
            quot!_A(x1,x2,x3) = max{3, x1 - 158, x2 - 158, x3 - 1}
        
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = [1]
      pi(p) = [1]
      pi(times) = [1]
      pi(|0|) = []
      pi(div) = [2]
      pi(quot) = [3]

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

  r22

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              eq#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 5
              pr_A(x1,x2) = 15
              |0|_A() = 7
              true_A() = 15
              eq#!_A(x1,x2) = x1 + x2 + 9
              s!_A(x1) = x1 + 8
              pr!_A(x1,x2) = 9
              |0|!_A() = 7
              true!_A() = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = 0
            s_A(x1) = 0
            pr_A(x1,x2) = max{4, x1 - 8, x2 - 16}
            |0|_A = 7
            true_A = 4
            eq#!_A(x1,x2) = 0
            s!_A(x1) = x1
            pr!_A(x1,x2) = max{2, x2 - 16}
            |0|!_A = 0
            true!_A = 1
        
  
    partial status:
  
      pi(eq#) = [1]
      pi(s) = [1]
      pi(pr) = [1]
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.