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, r8, r11, r17, r18, r19, r20, r21

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:
              pr#_A(x1,x2) = x1
              s_A(x1) = x1 + 20
              divides#_A(x1,x2) = x2
              div#_A(x1,x2) = x1
              div_A(x1,x2) = x1 + x2 + 1
              zero#_A(x1) = x1
              if#_A(x1,x2,x3) = x2
              eq_A(x1,x2) = 4
              |0|_A() = 5
              plus_A(x1,x2) = x1 + x2 + 1
              zero_A(x1) = 1
              false_A() = 4
              divides_A(x1,x2) = 4
              times_A(x1,x2) = x1 + 1
              quot#_A(x1,x2,x3) = x1
              true_A() = 4
              pr#!_A(x1,x2) = x1 + x2 + 3
              s!_A(x1) = 39
              divides#!_A(x1,x2) = x2 + 42
              div#!_A(x1,x2) = 41
              div!_A(x1,x2) = 42
              zero#!_A(x1) = 40
              if#!_A(x1,x2,x3) = x1 + x2 + x3
              eq!_A(x1,x2) = x1 + 36
              |0|!_A() = 19
              plus!_A(x1,x2) = x1 + 36
              zero!_A(x1) = 38
              false!_A() = 40
              divides!_A(x1,x2) = x2 + 38
              times!_A(x1,x2) = x1 + 15
              quot#!_A(x1,x2,x3) = 41
              true!_A() = 18
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            pr#_A(x1,x2) = 1
            s_A(x1) = x1 + 21
            divides#_A(x1,x2) = 1
            div#_A(x1,x2) = 1
            div_A(x1,x2) = max{x1 + 14, x2 - 4}
            zero#_A(x1) = 1
            if#_A(x1,x2,x3) = 1
            eq_A(x1,x2) = 52
            |0|_A = 0
            plus_A(x1,x2) = max{x1 + 78, x2 + 5}
            zero_A(x1) = max{45, x1 - 27}
            false_A = 52
            divides_A(x1,x2) = max{52, x1 + 6}
            times_A(x1,x2) = 32
            quot#_A(x1,x2,x3) = 1
            true_A = 52
            pr#!_A(x1,x2) = x2 + 47
            s!_A(x1) = max{87, x1 - 16}
            divides#!_A(x1,x2) = max{x1 + 100, x2 + 100}
            div#!_A(x1,x2) = 99
            div!_A(x1,x2) = max{61, x1 - 45}
            zero#!_A(x1) = 7
            if#!_A(x1,x2,x3) = max{x1 - 46, x3 + 67}
            eq!_A(x1,x2) = 37
            |0|!_A = 39
            plus!_A(x1,x2) = max{x1 - 38, x2 + 41}
            zero!_A(x1) = 62
            false!_A = 36
            divides!_A(x1,x2) = max{90, x1 - 19, x2 - 46}
            times!_A(x1,x2) = max{98, x1 + 77, x2 + 38}
            quot#!_A(x1,x2,x3) = 99
            true!_A = 38
        
  
    partial status:
  
      pi(pr#) = []
      pi(s) = []
      pi(divides#) = []
      pi(div#) = []
      pi(div) = [1, 2]
      pi(zero#) = []
      pi(if#) = []
      pi(eq) = []
      pi(|0|) = []
      pi(plus) = [1, 2]
      pi(zero) = []
      pi(false) = []
      pi(divides) = [2]
      pi(times) = []
      pi(quot#) = []
      pi(true) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6



-- SCC decomposition.

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))
p4: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- 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#(s(x),s(y),z) -> quot#(x,y,z)
p4: 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

  r9, 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!):
        
          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 + 6
              times_A(x1,x2) = x2 + 7
              zero_A(x1) = 4
              quot#_A(x1,x2,x3) = x1
              s_A(x1) = x1 + 4
              |0|_A() = 12
              div#!_A(x1,x2) = x1 + 6
              div!_A(x1,x2) = x1 + 5
              times!_A(x1,x2) = x2 + 8
              zero!_A(x1) = 10
              quot#!_A(x1,x2,x3) = x1 + 6
              s!_A(x1) = 9
              |0|!_A() = 7
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{1, x1 - 309}
            div_A(x1,x2) = max{308, x1 + 1, x2 - 2}
            times_A(x1,x2) = max{x1 - 113, x2 + 195}
            zero_A(x1) = max{113, x1 - 125}
            quot#_A(x1,x2,x3) = max{1, x1 - 309}
            s_A(x1) = max{83, x1}
            |0|_A = 194
            div#!_A(x1,x2) = x1 + 7
            div!_A(x1,x2) = max{312, x1 + 3}
            times!_A(x1,x2) = max{316, x1 - 309, x2 + 239}
            zero!_A(x1) = max{312, x1 - 211}
            quot#!_A(x1,x2,x3) = x1 + 7
            s!_A(x1) = max{10, x1 + 8}
            |0|!_A = 9
        
  
    partial status:
  
      pi(div#) = [1]
      pi(div) = [1, 2]
      pi(times) = [1, 2]
      pi(zero) = [1]
      pi(quot#) = [1]
      pi(s) = [1]
      pi(|0|) = []

The following rules are removed:

  p1, p3



-- 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

  (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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              div#_A(x1,x2) = x2
              quot#_A(x1,x2,x3) = x2
              |0|_A() = 5
              s_A(x1) = 3
              div#!_A(x1,x2) = 10
              quot#!_A(x1,x2,x3) = 10
              |0|!_A() = 6
              s!_A(x1) = x1 + 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = max{93, x1 - 1, x2 - 101}
            quot#_A(x1,x2,x3) = max{93, x1 - 1, x2 - 217, x3 - 101}
            |0|_A = 216
            s_A(x1) = 193
            div#!_A(x1,x2) = max{20, x2 - 194}
            quot#!_A(x1,x2,x3) = max{1, x2 - 195}
            |0|!_A = 86
            s!_A(x1) = max{88, x1 + 19}
        
  
    partial status:
  
      pi(div#) = [1]
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = [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: 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

  (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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              times#_A(x1,x2) = 0
              s_A(x1) = 8
              times#!_A(x1,x2) = 0
              s!_A(x1) = x1 + 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = x1
            s_A(x1) = x1 + 18
            times#!_A(x1,x2) = max{10, x1 - 1}
            s!_A(x1) = x1 + 20
        
  
    partial status:
  
      pi(times#) = [2]
      pi(s) = [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: 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:
  
      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 + x2
              s_A(x1) = x1 + 4
              p_A(x1) = x1
              plus#!_A(x1,x2) = 13
              s!_A(x1) = 14
              p!_A(x1) = 13
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 105, x2 - 133}
            s_A(x1) = x1 + 104
            p_A(x1) = max{1, x1 - 103}
            plus#!_A(x1,x2) = x1 + 26
            s!_A(x1) = x1 + 25
            p!_A(x1) = max{5, x1 - 97}
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(p) = [1]

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, 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!):
        
          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
              |0|_A() = 11
              div_A(x1,x2) = x1 + x2 + 5
              plus#!_A(x1,x2) = 9
              s!_A(x1) = 10
              p!_A(x1) = 11
              |0|!_A() = 6
              div!_A(x1,x2) = x1 + 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 1}
            s_A(x1) = x1 + 5
            p_A(x1) = max{1, x1 - 1}
            |0|_A = 0
            div_A(x1,x2) = max{1, x1 - 63, x2 - 71}
            plus#!_A(x1,x2) = max{1, x2 - 2}
            s!_A(x1) = x1 + 39
            p!_A(x1) = max{47, x1 - 68}
            |0|!_A = 46
            div!_A(x1,x2) = 46
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(p) = [1]
      pi(|0|) = []
      pi(div) = [1, 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(|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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              eq#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 7
              eq#!_A(x1,x2) = x1 + 21
              s!_A(x1) = x1 + 20
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = max{x1, x2}
            s_A(x1) = x1 + 184
            eq#!_A(x1,x2) = max{7, x1 - 185, x2 - 185}
            s!_A(x1) = 6
        
  
    partial status:
  
      pi(eq#) = [1, 2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.