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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = max{x1 + 59, x2 + 48}
          s_A(x1) = max{47, x1 + 21}
          minus_A(x1,x2) = max{26, x1}
          m_A(x1,x2) = max{x1 + 5, x2 + 5}
          n_A(x1,x2) = x1 + 6
          k_A(x1,x2) = x1 + 4
          |0|_A = 1
    
      precedence:
      
        g# = m = n = k > s = minus > |0|
      
      partial status:
    
        pi(g#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(m) = [1, 2]
        pi(n) = [1]
        pi(k) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = 21
          s_A(x1) = 28
          minus_A(x1,x2) = max{2, x1}
          m_A(x1,x2) = max{17, x2 + 14}
          n_A(x1,x2) = max{8, x1 + 5}
          k_A(x1,x2) = 0
          |0|_A = 4
    
      precedence:
      
        minus > |0| > g# = s = m = n = k
      
      partial status:
    
        pi(g#) = []
        pi(s) = []
        pi(minus) = []
        pi(m) = [2]
        pi(n) = [1]
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = max{x1 + 15, x2 + 11}
          s_A(x1) = max{20, x1 + 14}
          minus_A(x1,x2) = max{15, x1 + 2}
          m_A(x1,x2) = max{x1 + 7, x2 + 7}
          n_A(x1,x2) = x1 + 3
          |0|_A = 2
    
      precedence:
      
        minus = m = n > g# = s = |0|
      
      partial status:
    
        pi(g#) = [1, 2]
        pi(s) = [1]
        pi(minus) = []
        pi(m) = [1, 2]
        pi(n) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = max{0, x1 - 2, x2 - 1}
          s_A(x1) = max{14, x1 + 11}
          minus_A(x1,x2) = 16
          m_A(x1,x2) = max{8, x1 - 12, x2 + 4}
          n_A(x1,x2) = max{24, x1 + 10}
          |0|_A = 9
    
      precedence:
      
        g# = s = minus = m = n = |0|
      
      partial status:
    
        pi(g#) = []
        pi(s) = [1]
        pi(minus) = []
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 4
          s_A(x1) = x1 + 4
          h#_A(x1) = x1 + 3
    
      precedence:
      
        f# = s = h#
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(h#) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{3, x1 + 2}
          s_A(x1) = x1 + 1
          h#_A(x1) = x1 + 2
    
      precedence:
      
        f# = s = h#
      
      partial status:
    
        pi(f#) = []
        pi(s) = [1]
        pi(h#) = [1]
    

The next rules are strictly ordered:

  p1, p2

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: 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          k#_A(x1,x2) = max{3, x1 - 2}
          s_A(x1) = x1 + 9
          minus_A(x1,x2) = max{4, x1 + 1}
          |0|_A = 0
    
      precedence:
      
        k# = minus = |0| > s
      
      partial status:
    
        pi(k#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          k#_A(x1,x2) = 2
          s_A(x1) = 1
          minus_A(x1,x2) = 3
          |0|_A = 0
    
      precedence:
      
        minus > k# > s = |0|
      
      partial status:
    
        pi(k#) = []
        pi(s) = []
        pi(minus) = []
        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:

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

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:

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

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:

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

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1,x2) = max{4, x1, x2 - 12}
          s_A(x1) = max{47, x1 + 39}
          if_A(x1,x2,x3) = max{x1 + 18, x2, x3}
          gt_A(x1,x2) = max{22, x2 + 2}
          not_A(x1) = max{34, x1 + 18}
          id_A(x1) = max{47, x1 + 38}
          true_A = 33
          false_A = 1
          |0|_A = 34
    
      precedence:
      
        s = true > p# = false > gt = not > if = id = |0|
      
      partial status:
    
        pi(p#) = [1]
        pi(s) = [1]
        pi(if) = [1, 2, 3]
        pi(gt) = [2]
        pi(not) = [1]
        pi(id) = [1]
        pi(true) = []
        pi(false) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1,x2) = max{87, x1 + 80}
          s_A(x1) = x1 + 47
          if_A(x1,x2,x3) = 6
          gt_A(x1,x2) = x2 + 15
          not_A(x1) = x1 + 15
          id_A(x1) = x1 + 46
          true_A = 3
          false_A = 14
          |0|_A = 0
    
      precedence:
      
        p# = s = if = gt = not = id = true = false > |0|
      
      partial status:
    
        pi(p#) = [1]
        pi(s) = [1]
        pi(if) = []
        pi(gt) = [2]
        pi(not) = [1]
        pi(id) = []
        pi(true) = []
        pi(false) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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:

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

The next rules are strictly ordered:

  p1

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