YES

We show the termination of the TRS R:

  g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
  n(|0|(),y) -> |0|()
  n(x,|0|()) -> |0|()
  n(s(x),s(y)) -> s(n(x,y))
  m(|0|(),y) -> y
  m(x,|0|()) -> x
  m(s(x),s(y)) -> s(m(x,y))
  k(|0|(),s(y)) -> |0|()
  k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
  t(x) -> p(x,x)
  p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
  p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
  p(|0|(),y) -> y
  p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  id(x) -> x
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  not(x) -> if(x,false(),true())
  and(x,false()) -> false()
  and(true(),true()) -> true()
  f(|0|()) -> true()
  f(s(x)) -> h(x)
  h(|0|()) -> false()
  h(s(x)) -> f(x)
  gt(s(x),|0|()) -> true()
  gt(|0|(),y) -> false()
  gt(s(x),s(y)) -> gt(x,y)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(s(x),s(y)) -> if#(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
p2: g#(s(x),s(y)) -> and#(f(s(x)),f(s(y)))
p3: g#(s(x),s(y)) -> f#(s(x))
p4: g#(s(x),s(y)) -> f#(s(y))
p5: g#(s(x),s(y)) -> t#(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|())))))
p6: g#(s(x),s(y)) -> g#(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))
p7: g#(s(x),s(y)) -> k#(minus(m(x,y),n(x,y)),s(s(|0|())))
p8: g#(s(x),s(y)) -> minus#(m(x,y),n(x,y))
p9: g#(s(x),s(y)) -> m#(x,y)
p10: g#(s(x),s(y)) -> n#(x,y)
p11: g#(s(x),s(y)) -> k#(n(s(x),s(y)),s(s(|0|())))
p12: g#(s(x),s(y)) -> n#(s(x),s(y))
p13: g#(s(x),s(y)) -> g#(minus(m(x,y),n(x,y)),n(s(x),s(y)))
p14: n#(s(x),s(y)) -> n#(x,y)
p15: m#(s(x),s(y)) -> m#(x,y)
p16: k#(s(x),s(y)) -> k#(minus(x,y),s(y))
p17: k#(s(x),s(y)) -> minus#(x,y)
p18: t#(x) -> p#(x,x)
p19: p#(s(x),s(y)) -> p#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p20: p#(s(x),s(y)) -> if#(gt(x,y),x,y)
p21: p#(s(x),s(y)) -> gt#(x,y)
p22: p#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p23: p#(s(x),s(y)) -> not#(gt(x,y))
p24: p#(s(x),s(y)) -> id#(x)
p25: p#(s(x),s(y)) -> id#(y)
p26: p#(s(x),x) -> p#(if(gt(x,x),id(x),id(x)),s(x))
p27: p#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p28: p#(s(x),x) -> gt#(x,x)
p29: p#(s(x),x) -> id#(x)
p30: p#(id(x),s(y)) -> p#(x,if(gt(s(y),y),y,s(y)))
p31: p#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p32: p#(id(x),s(y)) -> gt#(s(y),y)
p33: minus#(s(x),s(y)) -> minus#(x,y)
p34: not#(x) -> if#(x,false(),true())
p35: f#(s(x)) -> h#(x)
p36: h#(s(x)) -> f#(x)
p37: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p6, p13}
  {p35, p36}
  {p16}
  {p33}
  {p15}
  {p14}
  {p19, p26, p30}
  {p37}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(s(x),s(y)) -> g#(minus(m(x,y),n(x,y)),n(s(x),s(y)))
p2: g#(s(x),s(y)) -> g#(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r2, r3, r4, r5, r6, r7, r8, r9, r15, r16

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1,x2) = max{10, x1 + 7, x2 + 3}
        s_A(x1) = x1 + 14
        minus_A(x1,x2) = x1
        m_A(x1,x2) = max{x1 + 11, x2 + 7}
        n_A(x1,x2) = x1
        k_A(x1,x2) = x1 + 2
        |0|_A = 0
  
    precedence:
    
      minus = n = |0| > g# = s = m = k
    
    partial status:
  
      pi(g#) = []
      pi(s) = []
      pi(minus) = []
      pi(m) = [2]
      pi(n) = []
      pi(k) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(s(x),s(y)) -> g#(minus(m(x,y),n(x,y)),n(s(x),s(y)))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(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: g#(s(x),s(y)) -> g#(minus(m(x,y),n(x,y)),n(s(x),s(y)))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r2, r3, r4, r5, r6, r7, r15, r16

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1,x2) = max{1, x1, x2 - 6}
        s_A(x1) = max{17, x1 + 15}
        minus_A(x1,x2) = max{x1 + 1, x2 + 3}
        m_A(x1,x2) = max{x1 + 13, x2 + 7}
        n_A(x1,x2) = max{5, x1 + 3}
        |0|_A = 4
  
    precedence:
    
      g# = s = minus = m = n = |0|
    
    partial status:
  
      pi(g#) = [1]
      pi(s) = []
      pi(minus) = [2]
      pi(m) = [2]
      pi(n) = [1]
      pi(|0|) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(s(x)) -> h#(x)
p2: h#(s(x)) -> f#(x)

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1 + 2
        s_A(x1) = x1 + 1
        h#_A(x1) = x1 + 1
  
    precedence:
    
      s = h# > f#
    
    partial status:
  
      pi(f#) = []
      pi(s) = [1]
      pi(h#) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(s(x)) -> h#(x)

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: k#(s(x),s(y)) -> k#(minus(x,y),s(y))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r15, r16

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        k#_A(x1,x2) = x1 + 2
        s_A(x1) = x1 + 3
        minus_A(x1,x2) = x1 + 2
        |0|_A = 0
  
    precedence:
    
      s > k# = minus = |0|
    
    partial status:
  
      pi(k#) = [1]
      pi(s) = []
      pi(minus) = [1]
      pi(|0|) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      minus# = s
    
    partial status:
  
      pi(minus#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: m#(s(x),s(y)) -> m#(x,y)

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        m#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      m# = s
    
    partial status:
  
      pi(m#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: n#(s(x),s(y)) -> n#(x,y)

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        n#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      n# = s
    
    partial status:
  
      pi(n#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: p#(s(x),s(y)) -> p#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p2: p#(id(x),s(y)) -> p#(x,if(gt(s(y),y),y,s(y)))
p3: p#(s(x),x) -> p#(if(gt(x,x),id(x),id(x)),s(x))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r17, r18, r19, r20, r27, r28, r29

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        p#_A(x1,x2) = max{17, x1 - 5, x2 - 26}
        s_A(x1) = x1 + 40
        if_A(x1,x2,x3) = max{39, x2 + 15, x3}
        gt_A(x1,x2) = max{x1 - 41, x2 + 41}
        not_A(x1) = x1 + 77
        id_A(x1) = x1 + 19
        true_A = 37
        false_A = 39
        |0|_A = 36
  
    precedence:
    
      p# = s = if = gt = not > id > true = false = |0|
    
    partial status:
  
      pi(p#) = []
      pi(s) = [1]
      pi(if) = [3]
      pi(gt) = [2]
      pi(not) = [1]
      pi(id) = []
      pi(true) = []
      pi(false) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: p#(s(x),s(y)) -> p#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p2: p#(id(x),s(y)) -> p#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(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: p#(s(x),s(y)) -> p#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p2: p#(id(x),s(y)) -> p#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r17, r18, r19, r20, r27, r28, r29

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        p#_A(x1,x2) = max{12, x1 + 5, x2 - 2}
        s_A(x1) = x1 + 40
        if_A(x1,x2,x3) = max{13, x1 + 5, x2 + 1, x3}
        gt_A(x1,x2) = max{x1 - 41, x2 + 16}
        not_A(x1) = x1 + 17
        id_A(x1) = x1 + 38
        true_A = 16
        false_A = 15
        |0|_A = 1
  
    precedence:
    
      true > p# = false > s = if = gt = not = id = |0|
    
    partial status:
  
      pi(p#) = []
      pi(s) = [1]
      pi(if) = [1, 2, 3]
      pi(gt) = [2]
      pi(not) = [1]
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: p#(id(x),s(y)) -> p#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(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: p#(id(x),s(y)) -> p#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r18, r19, r27, r28, r29

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        p#_A(x1,x2) = max{0, x1 - 1}
        id_A(x1) = x1 + 17
        s_A(x1) = max{17, x1 + 6}
        if_A(x1,x2,x3) = max{x1 + 11, x2 + 7, x3 + 1}
        gt_A(x1,x2) = 18
        |0|_A = 0
        false_A = 1
        true_A = 5
  
    precedence:
    
      p# > id = s = if = gt = |0| = false = true
    
    partial status:
  
      pi(p#) = []
      pi(id) = [1]
      pi(s) = [1]
      pi(if) = []
      pi(gt) = []
      pi(|0|) = []
      pi(false) = []
      pi(true) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: g(s(x),s(y)) -> if(and(f(s(x)),f(s(y))),t(g(k(minus(m(x,y),n(x,y)),s(s(|0|()))),k(n(s(x),s(y)),s(s(|0|()))))),g(minus(m(x,y),n(x,y)),n(s(x),s(y))))
r2: n(|0|(),y) -> |0|()
r3: n(x,|0|()) -> |0|()
r4: n(s(x),s(y)) -> s(n(x,y))
r5: m(|0|(),y) -> y
r6: m(x,|0|()) -> x
r7: m(s(x),s(y)) -> s(m(x,y))
r8: k(|0|(),s(y)) -> |0|()
r9: k(s(x),s(y)) -> s(k(minus(x,y),s(y)))
r10: t(x) -> p(x,x)
r11: p(s(x),s(y)) -> s(s(p(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r12: p(s(x),x) -> p(if(gt(x,x),id(x),id(x)),s(x))
r13: p(|0|(),y) -> y
r14: p(id(x),s(y)) -> s(p(x,if(gt(s(y),y),y,s(y))))
r15: minus(x,|0|()) -> x
r16: minus(s(x),s(y)) -> minus(x,y)
r17: id(x) -> x
r18: if(true(),x,y) -> x
r19: if(false(),x,y) -> y
r20: not(x) -> if(x,false(),true())
r21: and(x,false()) -> false()
r22: and(true(),true()) -> true()
r23: f(|0|()) -> true()
r24: f(s(x)) -> h(x)
r25: h(|0|()) -> false()
r26: h(s(x)) -> f(x)
r27: gt(s(x),|0|()) -> true()
r28: gt(|0|(),y) -> false()
r29: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        gt#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      gt# = s
    
    partial status:
  
      pi(gt#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.