YES
We show the termination of the TRS R:

  plus(x,|0|()) -> x
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,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: times#(s(x),y) -> plus#(y,times(x,y))
p3: times#(s(x),y) -> times#(x,y)
p4: div#(x,y) -> quot#(x,y,y)
p5: quot#(s(x),s(y),z) -> quot#(x,y,z)
p6: quot#(x,|0|(),s(z)) -> div#(x,s(z))
p7: div#(div(x,y),z) -> div#(x,times(y,z))
p8: div#(div(x,y),z) -> times#(y,z)
p9: eq#(s(x),s(y)) -> eq#(x,y)
p10: divides#(y,x) -> eq#(x,times(div(x,y),y))
p11: divides#(y,x) -> times#(div(x,y),y)
p12: divides#(y,x) -> div#(x,y)
p13: prime#(s(s(x))) -> pr#(s(s(x)),s(x))
p14: pr#(x,s(s(y))) -> if#(divides(s(s(y)),x),x,s(y))
p15: pr#(x,s(s(y))) -> divides#(s(s(y)),x)
p16: if#(false(),x,y) -> pr#(x,y)

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The estimated dependency graph contains the following SCCs:

  {p14, p16}
  {p4, p5, p6, p7}
  {p3}
  {p1}
  {p9}


-- 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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r1, r2, r5, r7, r13, r14, r15, r16, r17

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:
          if#_A(x1,x2,x3) = x2 + 14
          false_A() = 10
          pr#_A(x1,x2) = x1 + 14
          s_A(x1) = x1 + 6
          divides_A(x1,x2) = x2 + 13
          plus_A(x1,x2) = x1 + x2 + 7
          |0|_A() = 1
          times_A(x1,x2) = x1 + x2 + 7
          div_A(x1,x2) = x1 + x2 + 14
          eq_A(x1,x2) = x1 + 10
          true_A() = 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 4, x3 - 22}
          false_A = 1
          pr#_A(x1,x2) = max{0, x1 - 4, x2 - 27}
          s_A(x1) = max{19, x1 + 5}
          divides_A(x1,x2) = max{2, x1 - 25, x2 - 1}
          plus_A(x1,x2) = max{18, x1 + 2, x2 + 6}
          |0|_A = 7
          times_A(x1,x2) = max{6, x1 - 20, x2 + 1}
          div_A(x1,x2) = max{25, x1 - 45, x2 - 25}
          eq_A(x1,x2) = max{2, x1 - 20}
          true_A = 2
          if#!_A(x1,x2,x3) = max{15, x3 - 8}
          false!_A = 4
          pr#!_A(x1,x2) = max{12, x2 - 9}
          s!_A(x1) = max{0, x1 - 76}
          divides!_A(x1,x2) = max{11, x1 - 11, x2 + 6}
          plus!_A(x1,x2) = max{13, x1 - 78, x2 - 59}
          |0|!_A = 1
          times!_A(x1,x2) = max{12, x1 - 57, x2 - 79}
          div!_A(x1,x2) = max{13, x1 - 26, x2 - 78}
          eq!_A(x1,x2) = max{5, x1 - 20}
          true!_A = 3
  
    partial status:
  
      pi(if#) = []
      pi(false) = []
      pi(pr#) = [1]
      pi(s) = []
      pi(divides) = [2]
      pi(plus) = [2]
      pi(|0|) = []
      pi(times) = []
      pi(div) = []
      pi(eq) = []
      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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: 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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          div#_A(x1,x2) = x1 + 4
          div_A(x1,x2) = x1 + 5
          times_A(x1,x2) = x1 + 11
          quot#_A(x1,x2,x3) = x1 + 4
          |0|_A() = 3
          s_A(x1) = x1 + 12
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{1, x1 - 4}
          div_A(x1,x2) = max{3, x1 + 2, x2 - 22}
          times_A(x1,x2) = max{3, x1 - 27, x2 - 27}
          quot#_A(x1,x2,x3) = max{1, x1 - 4}
          |0|_A = 5
          s_A(x1) = max{19, x1 + 2}
          div#!_A(x1,x2) = 0
          div!_A(x1,x2) = max{7, x1 - 6, x2 - 10}
          times!_A(x1,x2) = max{x1 - 20, x2 + 9}
          quot#!_A(x1,x2,x3) = 0
          |0|!_A = 4
          s!_A(x1) = 5
  
    partial status:
  
      pi(div#) = []
      pi(div) = [1]
      pi(times) = [1]
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = [1]

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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: 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: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r4, r14, r15, 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:
          div#_A(x1,x2) = x2 + 1
          quot#_A(x1,x2,x3) = 1
          |0|_A() = 3
          s_A(x1) = 0
          times_A(x1,x2) = 4
          eq_A(x1,x2) = x2 + 2
          false_A() = 1
          pr_A(x1,x2) = x1 + 8
          true_A() = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{80, x1 - 1, x2 + 3}
          quot#_A(x1,x2,x3) = max{80, x1 - 1, x2 - 82, x3 + 3}
          |0|_A = 81
          s_A(x1) = 76
          times_A(x1,x2) = max{32, x1 + 2, x2 - 3}
          eq_A(x1,x2) = max{10, x1 - 60, x2 - 82}
          false_A = 11
          pr_A(x1,x2) = max{81, x1 - 74}
          true_A = 22
          div#!_A(x1,x2) = x2 + 3
          quot#!_A(x1,x2,x3) = max{2, x2 - 1}
          |0|!_A = 42
          s!_A(x1) = max{81, x1 - 81}
          times!_A(x1,x2) = max{84, x1 - 40}
          eq!_A(x1,x2) = max{86, x1 + 2, x2 + 2}
          false!_A = 84
          pr!_A(x1,x2) = max{86, x1 - 74, x2 - 74}
          true!_A = 85
  
    partial status:
  
      pi(div#) = [2]
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = []
      pi(times) = []
      pi(eq) = [2]
      pi(false) = []
      pi(pr) = [1]
      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: times#(s(x),y) -> times#(x,y)

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r4

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 + 13
          times_A(x1,x2) = x1 + 11
          |0|_A() = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times#_A(x1,x2) = 0
          s_A(x1) = max{48, x1 - 16}
          times_A(x1,x2) = max{66, x1 - 51, x2 - 51}
          |0|_A = 17
          times#!_A(x1,x2) = max{1, x1 - 137, x2 - 1}
          s!_A(x1) = max{3, x1 - 136}
          times!_A(x1,x2) = max{6, x1 - 137, x2 - 1}
          |0|!_A = 4
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = [1]
      pi(times) = [1]
      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)

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: if(false(),x,y) -> pr(x,y)

The set of usable rules consists of

  r4

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
          s_A(x1) = x1 + 5
          times_A(x1,x2) = x1 + 3
          |0|_A() = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = 0
          s_A(x1) = max{45, x1 - 23}
          times_A(x1,x2) = max{47, x1 - 47, x2 - 69}
          |0|_A = 20
          plus#!_A(x1,x2) = max{1, x1 - 112, x2 - 1}
          s!_A(x1) = max{25, x1 - 111}
          times!_A(x1,x2) = max{28, x1 - 193, x2 - 194}
          |0|!_A = 26
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(times) = [1]
      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: eq#(s(x),s(y)) -> eq#(x,y)

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(|0|(),y) -> y
r3: plus(s(x),y) -> s(plus(x,y))
r4: times(|0|(),y) -> |0|()
r5: times(s(|0|()),y) -> y
r6: times(s(x),y) -> plus(y,times(x,y))
r7: div(|0|(),y) -> |0|()
r8: div(x,y) -> quot(x,y,y)
r9: quot(|0|(),s(y),z) -> |0|()
r10: quot(s(x),s(y),z) -> quot(x,y,z)
r11: quot(x,|0|(),s(z)) -> s(div(x,s(z)))
r12: div(div(x,y),z) -> div(x,times(y,z))
r13: eq(|0|(),|0|()) -> true()
r14: eq(s(x),|0|()) -> false()
r15: eq(|0|(),s(y)) -> false()
r16: eq(s(x),s(y)) -> eq(x,y)
r17: divides(y,x) -> eq(x,times(div(x,y),y))
r18: prime(s(s(x))) -> pr(s(s(x)),s(x))
r19: pr(x,s(|0|())) -> true()
r20: pr(x,s(s(y))) -> if(divides(s(s(y)),x),x,s(y))
r21: if(true(),x,y) -> false()
r22: 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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          eq#_A(x1,x2) = x1 + x2 + 5
          s_A(x1) = x1 + 4
          div_A(x1,x2) = x1
          |0|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          eq#_A(x1,x2) = 0
          s_A(x1) = max{8, x1 - 23}
          div_A(x1,x2) = max{21, x1 - 22, x2 - 22}
          |0|_A = 18
          eq#!_A(x1,x2) = max{12, x2 - 9}
          s!_A(x1) = max{11, x1 + 5}
          div!_A(x1,x2) = max{x1 + 24, x2 + 22}
          |0|!_A = 10
  
    partial status:
  
      pi(eq#) = []
      pi(s) = [1]
      pi(div) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.