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

  r4, r8, r11, r13, r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          pr#_A(x1,x2) = x2
          s_A(x1) = x1 + 3
          divides#_A(x1,x2) = 6
          div#_A(x1,x2) = 6
          div_A(x1,x2) = x1 + 10
          zero#_A(x1) = 6
          if#_A(x1,x2,x3) = x3 + 1
          eq_A(x1,x2) = x1 + 8
          |0|_A() = 8
          plus_A(x1,x2) = x2 + 1
          zero_A(x1) = 1
          false_A() = 6
          divides_A(x1,x2) = x2 + 7
          times_A(x1,x2) = x1 + 6
          quot#_A(x1,x2,x3) = 6
          quot_A(x1,x2,x3) = 9
          true_A() = 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          pr#_A(x1,x2) = 1
          s_A(x1) = 0
          divides#_A(x1,x2) = max{1, x1 - 1}
          div#_A(x1,x2) = 1
          div_A(x1,x2) = max{2, x1 - 12, x2 - 12}
          zero#_A(x1) = 1
          if#_A(x1,x2,x3) = 1
          eq_A(x1,x2) = 16
          |0|_A = 0
          plus_A(x1,x2) = max{4, x1, x2}
          zero_A(x1) = x1
          false_A = 5
          divides_A(x1,x2) = 15
          times_A(x1,x2) = max{3, x2 - 1}
          quot#_A(x1,x2,x3) = 1
          quot_A(x1,x2,x3) = max{3, x1 - 1, x2 - 1, x3 - 1}
          true_A = 3
          pr#!_A(x1,x2) = 5
          s!_A(x1) = 6
          divides#!_A(x1,x2) = x1 + 4
          div#!_A(x1,x2) = 3
          div!_A(x1,x2) = max{7, x1 - 10, x2 - 12}
          zero#!_A(x1) = 2
          if#!_A(x1,x2,x3) = x3 + 1
          eq!_A(x1,x2) = 14
          |0|!_A = 13
          plus!_A(x1,x2) = max{x1 + 11, x2 + 10}
          zero!_A(x1) = 8
          false!_A = 12
          divides!_A(x1,x2) = max{x1 + 13, x2 - 36}
          times!_A(x1,x2) = max{1, x1 - 23}
          quot#!_A(x1,x2,x3) = 3
          quot!_A(x1,x2,x3) = max{0, x1 - 11, x2 - 13}
          true!_A = 11
  
    partial status:
  
      pi(pr#) = []
      pi(s) = [1]
      pi(divides#) = []
      pi(div#) = []
      pi(div) = [1]
      pi(zero#) = []
      pi(if#) = []
      pi(eq) = [1]
      pi(|0|) = []
      pi(plus) = [2]
      pi(zero) = []
      pi(false) = []
      pi(divides) = []
      pi(times) = []
      pi(quot#) = []
      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

  r1, r8, r11

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          div#_A(x1,x2) = x1 + 6
          div_A(x1,x2) = x1 + x2 + 17
          times_A(x1,x2) = x1 + 6
          zero_A(x1) = 10
          quot#_A(x1,x2,x3) = x1 + 6
          s_A(x1) = x1 + 5
          |0|_A() = 4
          p_A(x1) = x1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{1, x1 - 187}
          div_A(x1,x2) = max{186, x1 + 3, x2 - 44}
          times_A(x1,x2) = max{43, x1 - 245, x2 - 245}
          zero_A(x1) = x1 + 6
          quot#_A(x1,x2,x3) = max{1, x1 - 187}
          s_A(x1) = max{26, x1 + 17}
          |0|_A = 5
          p_A(x1) = max{25, x1 - 6}
          div#!_A(x1,x2) = 0
          div!_A(x1,x2) = max{278, x1 + 94, x2 + 238}
          times!_A(x1,x2) = max{237, x2 - 1}
          zero!_A(x1) = max{279, x1 - 88}
          quot#!_A(x1,x2,x3) = 0
          s!_A(x1) = max{0, x1 - 64}
          |0|!_A = 0
          p!_A(x1) = max{0, x1 - 81}
  
    partial status:
  
      pi(div#) = []
      pi(div) = [1]
      pi(times) = []
      pi(zero) = []
      pi(quot#) = []
      pi(s) = [1]
      pi(|0|) = []
      pi(p) = []

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:
  
      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() = 0
          s_A(x1) = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{0, x1 - 2, x2 - 9}
          quot#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 81, x3 - 9}
          |0|_A = 79
          s_A(x1) = max{8, x1 + 2}
          div#!_A(x1,x2) = 158
          quot#!_A(x1,x2,x3) = 0
          |0|!_A = 159
          s!_A(x1) = 157
  
    partial status:
  
      pi(div#) = [2]
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []

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

  r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          times#_A(x1,x2) = x1
          s_A(x1) = x1 + 6
          eq_A(x1,x2) = x1 + 2
          |0|_A() = 0
          true_A() = 1
          false_A() = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times#_A(x1,x2) = 0
          s_A(x1) = 51
          eq_A(x1,x2) = max{47, x1 - 15, x2 - 15}
          |0|_A = 94
          true_A = 48
          false_A = 48
          times#!_A(x1,x2) = max{52, x1 - 15, x2 - 1}
          s!_A(x1) = x1 + 51
          eq!_A(x1,x2) = max{147, x2 - 38}
          |0|!_A = 44
          true!_A = 57
          false!_A = 45
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = []
      pi(eq) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1 + x2 + 10
          s_A(x1) = x1 + 9
          p_A(x1) = x1
          |0|_A() = 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 32, x2 - 32}
          s_A(x1) = max{31, x1 + 9}
          p_A(x1) = max{23, x1 - 4}
          |0|_A = 3
          plus#!_A(x1,x2) = max{38, x1 + 30}
          s!_A(x1) = max{37, x1 - 92}
          p!_A(x1) = 37
          |0|!_A = 36
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(p) = []
      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

  r1, r2, r23

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1 + 8
          s_A(x1) = x1 + 9
          p_A(x1) = x1
          |0|_A() = 3
          pr_A(x1,x2) = 3
          true_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 25, x2 - 47}
          s_A(x1) = x1 + 46
          p_A(x1) = max{25, x1 - 9}
          |0|_A = 4
          pr_A(x1,x2) = max{153, x1 - 253, x2 - 275}
          true_A = 2
          plus#!_A(x1,x2) = max{4, x2 - 41}
          s!_A(x1) = x1 + 7
          p!_A(x1) = 6
          |0|!_A = 5
          pr!_A(x1,x2) = max{55, x1 - 137, x2 - 275}
          true!_A = 1
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(p) = []
      pi(|0|) = []
      pi(pr) = []
      pi(true) = []

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

  r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          eq#_A(x1,x2) = x1 + x2 + 8
          s_A(x1) = x1 + 7
          times_A(x1,x2) = x1 + 1
          |0|_A() = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          eq#_A(x1,x2) = max{0, x1 - 17, x2 - 17}
          s_A(x1) = x1 + 16
          times_A(x1,x2) = max{60, x1 - 98, x2 - 213}
          |0|_A = 17
          eq#!_A(x1,x2) = max{12, x1 - 17, x2 - 17}
          s!_A(x1) = max{11, x1 + 6}
          times!_A(x1,x2) = max{5, x1 - 4}
          |0|!_A = 4
  
    partial status:
  
      pi(eq#) = [1]
      pi(s) = [1]
      pi(times) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.