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

  r9, r13, r14, r15, r16, r17

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{2, x1 - 12, x2 - 17, x3 - 33}
            false_A = 13
            pr#_A(x1,x2) = max{2, x1 - 17, x2 - 48}
            s_A(x1) = max{32, x1 + 15}
            divides_A(x1,x2) = max{13, x1 - 36}
            quot_A(x1,x2,x3) = max{36, x2 - 33, x3 - 33}
            |0|_A = 11
            eq_A(x1,x2) = 13
            true_A = 13
            times_A(x1,x2) = max{30, x2 + 11}
            div_A(x1,x2) = 21
            if#!_A(x1,x2,x3) = max{35, x1 - 12, x3 + 1}
            false!_A = 0
            pr#!_A(x1,x2) = max{34, x2 - 11}
            s!_A(x1) = 0
            divides!_A(x1,x2) = max{20, x1 - 49, x2 - 12}
            quot!_A(x1,x2,x3) = 0
            |0|!_A = 2
            eq!_A(x1,x2) = max{2, x1 - 12}
            true!_A = 1
            times!_A(x1,x2) = 0
            div!_A(x1,x2) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if#_A(x1,x2,x3) = max{42, x1 + 15, x2 + 41, x3 + 7}
            false_A = 26
            pr#_A(x1,x2) = max{42, x1 + 41, x2 - 8}
            s_A(x1) = 7
            divides_A(x1,x2) = max{x1 - 1, x2 + 26}
            quot_A(x1,x2,x3) = max{37, x1 - 1}
            |0|_A = 37
            eq_A(x1,x2) = 26
            true_A = 26
            times_A(x1,x2) = max{0, x1 - 44}
            div_A(x1,x2) = max{37, x1, x2}
            if#!_A(x1,x2,x3) = max{16, x1 - 7}
            false!_A = 13
            pr#!_A(x1,x2) = 18
            s!_A(x1) = 15
            divides!_A(x1,x2) = max{17, x1 + 12, x2 - 39}
            quot!_A(x1,x2,x3) = max{6, x2 - 8}
            |0|!_A = 16
            eq!_A(x1,x2) = max{16, x1 - 40}
            true!_A = 14
            times!_A(x1,x2) = 4
            div!_A(x1,x2) = x2 + 7
        
  
    partial status:
  
      pi(if#) = [3]
      pi(false) = []
      pi(pr#) = [2]
      pi(s) = [1]
      pi(divides) = [2]
      pi(quot) = [1]
      pi(|0|) = []
      pi(eq) = []
      pi(true) = []
      pi(times) = [1, 2]
      pi(div) = []

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

  r1, r2, r3, r4, r5, r6, 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:
          
            div#_A(x1,x2) = max{13, x2 - 13}
            div_A(x1,x2) = max{x1 + 14, x2 + 3}
            times_A(x1,x2) = max{26, x2}
            quot#_A(x1,x2,x3) = max{13, x3 - 14}
            |0|_A = 0
            s_A(x1) = 13
            plus_A(x1,x2) = max{26, x1, x2}
            div#!_A(x1,x2) = 0
            div!_A(x1,x2) = max{28, x1 + 23, x2 + 1}
            times!_A(x1,x2) = 36
            quot#!_A(x1,x2,x3) = 0
            |0|!_A = 35
            s!_A(x1) = 29
            plus!_A(x1,x2) = 30
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div#_A(x1,x2) = 1
            div_A(x1,x2) = max{x1 + 3, x2 + 3}
            times_A(x1,x2) = x2 + 9
            quot#_A(x1,x2,x3) = 1
            |0|_A = 9
            s_A(x1) = 0
            plus_A(x1,x2) = max{9, x1 + 1, x2}
            div#!_A(x1,x2) = x2 + 6
            div!_A(x1,x2) = max{23, x1 - 10, x2 - 17}
            times!_A(x1,x2) = max{24, x2 + 20}
            quot#!_A(x1,x2,x3) = max{x2 - 2, x3 + 3}
            |0|!_A = 21
            s!_A(x1) = 8
            plus!_A(x1,x2) = x2 + 10
        
  
    partial status:
  
      pi(div#) = [1]
      pi(div) = [1, 2]
      pi(times) = [1, 2]
      pi(quot#) = [1]
      pi(|0|) = []
      pi(s) = [1]
      pi(plus) = [1, 2]

The following rules are removed:

  p1, p2, p3, p4

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

  (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!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = 0
            s_A(x1) = max{4, x1 - 10}
            times#!_A(x1,x2) = max{0, x1 - 80}
            s!_A(x1) = max{4, x1 - 79}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = 0
            s_A(x1) = max{7, x1 - 2}
            times#!_A(x1,x2) = max{6, x1 - 8, x2 - 1}
            s!_A(x1) = max{5, x1 - 66}
        
  
    partial status:
  
      pi(times#) = [1]
      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)

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

  (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!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{7, x1 - 10}
            plus#!_A(x1,x2) = max{1, x1 - 57}
            s!_A(x1) = max{2, x1 - 56}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{11, x1 - 3}
            plus#!_A(x1,x2) = max{3, x1 - 12, x2 - 1}
            s!_A(x1) = max{2, x1 - 54}
        
  
    partial status:
  
      pi(plus#) = [1]
      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: 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

  (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!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = 0
            s_A(x1) = max{10, x1 - 16}
            eq#!_A(x1,x2) = 1
            s!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = 0
            s_A(x1) = max{33, x1 - 10}
            eq#!_A(x1,x2) = max{4, x1 - 139, x2 - 141}
            s!_A(x1) = max{3, x1 - 140}
        
  
    partial status:
  
      pi(eq#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.