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, r12, r16, r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if#_A(x1,x2,x3) = max{74, x3 + 13}
        false_A = 111
        pr#_A(x1,x2) = max{62, x2 + 12}
        s_A(x1) = max{61, x1 + 14}
        divides_A(x1,x2) = max{76, x1 - 93, x2 - 65}
        div_A(x1,x2) = max{7, x1 - 63, x2 + 2}
        |0|_A = 6
        quot_A(x1,x2,x3) = max{15, x1 - 62, x2 - 62, x3 - 62}
        eq_A(x1,x2) = max{112, x1 - 64, x2 + 2}
        true_A = 60
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if#_A(x1,x2,x3) = max{9, x3 - 9}
          false_A = 2
          pr#_A(x1,x2) = max{1, x2 - 27}
          s_A(x1) = x1 + 18
          divides_A(x1,x2) = x2 + 2
          div_A(x1,x2) = max{55, x1 + 1, x2}
          |0|_A = 19
          quot_A(x1,x2,x3) = max{x1 + 10, x2 - 20, x3 - 19}
          eq_A(x1,x2) = max{2, x1 - 20, x2 - 20}
          true_A = 2
          if#!_A(x1,x2,x3) = max{4, x2 - 57, x3 - 19}
          false!_A = 21
          pr#!_A(x1,x2) = max{4, x1 - 57, x2 - 37}
          s!_A(x1) = 3
          divides!_A(x1,x2) = max{0, x1 - 53, x2 - 57}
          div!_A(x1,x2) = max{5, x1 - 56, x2 + 4}
          |0|!_A = 1
          quot!_A(x1,x2,x3) = max{3, x1 - 20, x2 - 20, x3 - 19}
          eq!_A(x1,x2) = max{x1 - 20, x2 + 4}
          true!_A = 2
  
    partial status:
  
      pi(if#) = [3]
      pi(false) = []
      pi(pr#) = [2]
      pi(s) = []
      pi(divides) = []
      pi(div) = [2]
      pi(|0|) = []
      pi(quot) = []
      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: 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, r10, r12, r16, r17, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = 2
        div_A(x1,x2) = max{28, x1 - 40, x2 - 31}
        times_A(x1,x2) = max{20, x1 + 2}
        quot#_A(x1,x2,x3) = 2
        |0|_A = 23
        s_A(x1) = max{21, x1 + 3}
        quot_A(x1,x2,x3) = max{32, x1 - 24, x2 - 34, x3 - 22}
        eq_A(x1,x2) = max{28, x1 - 24, x2 - 31}
        true_A = 22
        false_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div#_A(x1,x2) = max{2, x1 + 1}
          div_A(x1,x2) = max{x1 + 3, x2 + 3}
          times_A(x1,x2) = max{0, x2 - 2}
          quot#_A(x1,x2,x3) = max{2, x1 + 1}
          |0|_A = 0
          s_A(x1) = x1 + 6
          quot_A(x1,x2,x3) = max{4, x1 - 1, x2 - 7, x3 - 7}
          eq_A(x1,x2) = max{5, x1 - 7, x2 - 3}
          true_A = 4
          false_A = 4
          div#!_A(x1,x2) = x1 + 18
          div!_A(x1,x2) = max{17, x1 - 4, x2}
          times!_A(x1,x2) = x1
          quot#!_A(x1,x2,x3) = x1 + 18
          |0|!_A = 15
          s!_A(x1) = x1 + 14
          quot!_A(x1,x2,x3) = max{16, x1 - 7, x2 - 7, x3 - 5}
          eq!_A(x1,x2) = max{24, x1 - 7, x2 - 3}
          true!_A = 13
          false!_A = 1
  
    partial status:
  
      pi(div#) = []
      pi(div) = []
      pi(times) = []
      pi(quot#) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(quot) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []

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

  r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div#_A(x1,x2) = x2 + 7
        quot#_A(x1,x2,x3) = max{6, x2 + 1}
        |0|_A = 16
        s_A(x1) = 7
        times_A(x1,x2) = max{17, x1 - 8, x2 - 4}
  
    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 - 18}
          quot#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 18, x3 - 18}
          |0|_A = 4
          s_A(x1) = max{17, x1 - 11}
          times_A(x1,x2) = max{6, x1 - 51, x2 - 94}
          div#!_A(x1,x2) = max{19, x1 + 1, x2 + 1}
          quot#!_A(x1,x2,x3) = max{x1 + 19, x2 + 16, x3 + 31}
          |0|!_A = 21
          s!_A(x1) = max{19, x1 - 149}
          times!_A(x1,x2) = max{23, x1 - 189, x2 - 137}
  
    partial status:
  
      pi(div#) = []
      pi(quot#) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(times) = []

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:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{x1 + 4, x2 + 5}
        s_A(x1) = x1 + 3
        times_A(x1,x2) = max{14, x1 - 14, x2 - 8}
        |0|_A = 13
  
    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{27, x1 - 16}
          times_A(x1,x2) = max{58, x1 - 29, x2 - 117}
          |0|_A = 10
          times#!_A(x1,x2) = 0
          s!_A(x1) = max{52, x1 - 148}
          times!_A(x1,x2) = max{100, x1 - 11, x2 - 29}
          |0|!_A = 53
  
    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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = 1
        s_A(x1) = max{4, x1 - 23}
        p_A(x1) = max{29, x1 + 24}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 7, x2 - 7}
          s_A(x1) = x1 + 6
          p_A(x1) = max{2, x1 - 3}
          plus#!_A(x1,x2) = max{1, x1 - 4}
          s!_A(x1) = max{39, x1 - 31}
          p!_A(x1) = max{40, x1 + 32}
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      pi(p) = []

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(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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{14, x1 - 1}
        s_A(x1) = max{13, x1 - 15}
        p_A(x1) = x1 + 18
        times_A(x1,x2) = max{5, x1 - 103, x2 - 46}
        |0|_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 2, x2 - 27}
          s_A(x1) = max{46, x1 + 21}
          p_A(x1) = max{28, x1 - 20}
          times_A(x1,x2) = max{79, x1 - 48, x2 - 6}
          |0|_A = 46
          plus#!_A(x1,x2) = max{1, x2 - 44}
          s!_A(x1) = max{9, x1 - 152}
          p!_A(x1) = max{9, x1 - 199}
          times!_A(x1,x2) = max{4, x1 - 347, x2 - 349}
          |0|!_A = 3
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      pi(p) = []
      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: 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

  r7, r16, r17, r18

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        eq#_A(x1,x2) = max{x1 + 6, x2 + 4}
        s_A(x1) = x1 + 5
        times_A(x1,x2) = max{11, x1 - 31, x2 - 19}
        |0|_A = 3
        eq_A(x1,x2) = max{x1 - 17, x2 + 7}
        true_A = 2
        false_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{56, x1 - 110}
          times_A(x1,x2) = max{48, x1 - 150, x2 - 220}
          |0|_A = 40
          eq_A(x1,x2) = max{7, x1 - 57, x2 - 41}
          true_A = 2
          false_A = 2
          eq#!_A(x1,x2) = x2 + 112
          s!_A(x1) = max{0, x1 - 107}
          times!_A(x1,x2) = max{3, x1 - 94, x2 - 274}
          |0|!_A = 2
          eq!_A(x1,x2) = max{2, x1 - 58, x2 - 106}
          true!_A = 1
          false!_A = 0
  
    partial status:
  
      pi(eq#) = []
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []
      pi(eq) = []
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.