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, r24

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{1, x2 - 2, x3 - 33}
            false_A = 1
            pr#_A(x1,x2) = max{1, x1 - 2, x2 - 49}
            s_A(x1) = max{32, x1 + 16}
            divides_A(x1,x2) = max{1, x1 - 31}
            div_A(x1,x2) = max{x1 + 7, x2 + 7}
            |0|_A = 32
            if_A(x1,x2,x3) = max{1, x1 - 18, x2 - 50, x3 - 67}
            true_A = 3
            if#!_A(x1,x2,x3) = max{21, x2 + 3, x3 - 17}
            false!_A = 0
            pr#!_A(x1,x2) = max{21, x1 + 3, x2 - 28}
            s!_A(x1) = 5
            divides!_A(x1,x2) = 2
            div!_A(x1,x2) = max{23, x2 - 9}
            |0|!_A = 3
            if!_A(x1,x2,x3) = max{0, x1 - 18, x3 - 33}
            true!_A = 0
        
        2. 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() = 7
              pr#_A(x1,x2) = x2
              s_A(x1) = x1 + 4
              divides_A(x1,x2) = 1
              div_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 3
              if_A(x1,x2,x3) = 7
              true_A() = 9
              if#!_A(x1,x2,x3) = x3 + 3
              false!_A() = 1
              pr#!_A(x1,x2) = x2
              s!_A(x1) = 6
              divides!_A(x1,x2) = 5
              div!_A(x1,x2) = x2 + 8
              |0|!_A() = 5
              if!_A(x1,x2,x3) = 3
              true!_A() = 2
        
  
    partial status:
  
      pi(if#) = []
      pi(false) = []
      pi(pr#) = []
      pi(s) = []
      pi(divides) = []
      pi(div) = [1, 2]
      pi(|0|) = []
      pi(if) = [1, 2]
      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: 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, r3, r8, r10

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 - 445}
            div_A(x1,x2) = max{444, x1 + 24, x2 - 111}
            times_A(x1,x2) = max{110, x1 + 25, x2 + 25}
            quot#_A(x1,x2,x3) = max{1, x1 - 445}
            |0|_A = 445
            s_A(x1) = max{292, x1 + 11}
            plus_A(x1,x2) = max{526, x1 + 244, x2 + 208}
            div#!_A(x1,x2) = 12
            div!_A(x1,x2) = max{x1 + 158, x2 + 22}
            times!_A(x1,x2) = max{603, x1 - 446, x2 + 21}
            quot#!_A(x1,x2,x3) = 12
            |0|!_A = 315
            s!_A(x1) = max{13, x1 + 8}
            plus!_A(x1,x2) = max{20, x1 - 11}
        
        2. 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) = x1 + x2 + 6
              quot#_A(x1,x2,x3) = x1
              |0|_A() = 4
              s_A(x1) = x1 + 3
              plus_A(x1,x2) = x1 + x2 + 4
              div#!_A(x1,x2) = x1 + 6
              div!_A(x1,x2) = x2 + 5
              times!_A(x1,x2) = 8
              quot#!_A(x1,x2,x3) = x1 + 6
              |0|!_A() = 11
              s!_A(x1) = 7
              plus!_A(x1,x2) = x1 + x2 + 3
        
  
    partial status:
  
      pi(div#) = []
      pi(div) = []
      pi(times) = [2]
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []
      pi(plus) = [1, 2]

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

  r10

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 - 2, x2 - 52}
            quot#_A(x1,x2,x3) = max{1, x1 - 2, x2 - 72}
            |0|_A = 70
            s_A(x1) = 51
            div_A(x1,x2) = max{59, x1 + 2, x2 - 24}
            div#!_A(x1,x2) = max{66, x2 + 13}
            quot#!_A(x1,x2,x3) = max{65, x2 - 1}
            |0|!_A = 67
            s!_A(x1) = x1 + 95
            div!_A(x1,x2) = max{x1 + 120, x2 + 72}
        
        2. 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) = x2
              quot#_A(x1,x2,x3) = x3
              |0|_A() = 3
              s_A(x1) = x1 + 1
              div_A(x1,x2) = x1 + x2 + 1
              div#!_A(x1,x2) = 1
              quot#!_A(x1,x2,x3) = 0
              |0|!_A() = 5
              s!_A(x1) = 4
              div!_A(x1,x2) = x2 + 6
        
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []
      pi(div) = [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!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = x1
            s_A(x1) = max{52, x1 + 15}
            times_A(x1,x2) = max{13, x1 - 70}
            |0|_A = 3
            times#!_A(x1,x2) = max{38, x1 - 14}
            s!_A(x1) = max{0, x1 - 18}
            times!_A(x1,x2) = max{129, x2 - 69}
            |0|!_A = 128
        
        2. 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 + 3
              |0|_A() = 6
              times#!_A(x1,x2) = x1
              s!_A(x1) = 6
              times!_A(x1,x2) = 9
              |0|!_A() = 7
        
  
    partial status:
  
      pi(times#) = [1]
      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, r7, 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!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 23, x2 - 23}
            s_A(x1) = x1 + 22
            p_A(x1) = max{2, x1}
            times_A(x1,x2) = max{0, x1 - 43}
            |0|_A = 0
            quot_A(x1,x2,x3) = max{1, x1 - 23, x2 - 23, x3 - 23}
            plus#!_A(x1,x2) = 24
            s!_A(x1) = x1 + 12
            p!_A(x1) = 23
            times!_A(x1,x2) = max{13, x1 - 1, x2 - 5}
            |0|!_A = 9
            quot!_A(x1,x2,x3) = max{13, x1 - 39, x2 - 23, x3 - 23}
        
        2. 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 + x2
              s_A(x1) = x1 + 1
              p_A(x1) = x1
              times_A(x1,x2) = x1 + 2
              |0|_A() = 6
              quot_A(x1,x2,x3) = x1 + 2
              plus#!_A(x1,x2) = x1 + x2 + 2
              s!_A(x1) = 4
              p!_A(x1) = x1 + 1
              times!_A(x1,x2) = x1 + 1
              |0|!_A() = 6
              quot!_A(x1,x2,x3) = x1 + x3 + 2
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(p) = [1]
      pi(times) = []
      pi(|0|) = []
      pi(quot) = [1, 2, 3]

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)))
p2: 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, p2}


-- Reduction pair.

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

p1: plus#(x,s(y)) -> plus#(x,p(s(y)))
p2: 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

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 - 8, x2 - 50}
            s_A(x1) = max{49, x1 + 9}
            p_A(x1) = max{43, x1 - 4}
            times_A(x1,x2) = max{6, x2 - 101}
            |0|_A = 0
            div_A(x1,x2) = max{0, x2 - 127}
            plus#!_A(x1,x2) = x1 + 79
            s!_A(x1) = x1 + 77
            p!_A(x1) = max{128, x1 - 50}
            times!_A(x1,x2) = max{81, x2 + 16}
            |0|!_A = 10
            div!_A(x1,x2) = max{11, x2 - 10}
        
        2. 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 + 1
              p_A(x1) = x1
              times_A(x1,x2) = x1 + 4
              |0|_A() = 5
              div_A(x1,x2) = x1 + x2 + 3
              plus#!_A(x1,x2) = 6
              s!_A(x1) = 7
              p!_A(x1) = 5
              times!_A(x1,x2) = x1 + x2 + 6
              |0|!_A() = 1
              div!_A(x1,x2) = 0
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(p) = [1]
      pi(times) = [1, 2]
      pi(|0|) = []
      pi(div) = []

The following rules are removed:

  p2



-- 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, r7, r10

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 - 115, x2 - 87}
            s_A(x1) = x1 + 86
            p_A(x1) = max{25, x1 - 55}
            times_A(x1,x2) = max{110, x1 + 103, x2 - 1}
            |0|_A = 2
            div_A(x1,x2) = max{9, x1 - 10}
            plus#!_A(x1,x2) = max{1, x2 - 26}
            s!_A(x1) = max{5, x1 - 31}
            p!_A(x1) = 6
            times!_A(x1,x2) = max{7, x1 - 8}
            |0|!_A = 3
            div!_A(x1,x2) = max{4, x1 - 26, x2 - 118}
        
        2. 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 + 1
              p_A(x1) = x1
              times_A(x1,x2) = x1 + 4
              |0|_A() = 5
              div_A(x1,x2) = x1 + x2 + 3
              plus#!_A(x1,x2) = x2 + 13
              s!_A(x1) = x1 + 12
              p!_A(x1) = x1 + 12
              times!_A(x1,x2) = 8
              |0|!_A() = 7
              div!_A(x1,x2) = x1 + 6
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(p) = [1]
      pi(times) = []
      pi(|0|) = []
      pi(div) = [2]

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

  r10

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) = 0
            s_A(x1) = 24
            div_A(x1,x2) = max{11, x1 + 8, x2 - 10}
            |0|_A = 26
            eq#!_A(x1,x2) = 0
            s!_A(x1) = x1 + 26
            div!_A(x1,x2) = x1 + 30
            |0|!_A = 25
        
        2. 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 + 4
              div_A(x1,x2) = x1 + x2 + 3
              |0|_A() = 5
              eq#!_A(x1,x2) = x1 + x2
              s!_A(x1) = 3
              div!_A(x1,x2) = x1 + x2 + 1
              |0|!_A() = 4
        
  
    partial status:
  
      pi(eq#) = [1]
      pi(s) = [1]
      pi(div) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.