YES
We show the termination of the TRS R:

  app(nil(),k) -> k
  app(l,nil()) -> l
  app(cons(x,l),k) -> cons(x,app(l,k))
  sum(cons(x,nil())) -> cons(x,nil())
  sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
  sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
  plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
  plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
  plus(zero(),y) -> y
  plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
  id(x) -> x
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  not(x) -> if(x,false(),true())
  gt(s(x),zero()) -> true()
  gt(zero(),y) -> false()
  gt(s(x),s(y)) -> gt(x,y)


-- SCC decomposition.

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

p1: app#(cons(x,l),k) -> app#(l,k)
p2: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))
p3: sum#(cons(x,cons(y,l))) -> plus#(x,y)
p4: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))
p5: sum#(app(l,cons(x,cons(y,k)))) -> app#(l,sum(cons(x,cons(y,k))))
p6: sum#(app(l,cons(x,cons(y,k)))) -> sum#(cons(x,cons(y,k)))
p7: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p8: plus#(s(x),s(y)) -> if#(gt(x,y),x,y)
p9: plus#(s(x),s(y)) -> gt#(x,y)
p10: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p11: plus#(s(x),s(y)) -> not#(gt(x,y))
p12: plus#(s(x),s(y)) -> id#(x)
p13: plus#(s(x),s(y)) -> id#(y)
p14: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))
p15: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p16: plus#(s(x),x) -> gt#(x,x)
p17: plus#(s(x),x) -> id#(x)
p18: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p19: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p20: plus#(id(x),s(y)) -> gt#(s(y),y)
p21: not#(x) -> if#(x,false(),true())
p22: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1}
  {p2}
  {p7, p14, p18}
  {p22}


-- Reduction pair.

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

p1: sum#(app(l,cons(x,cons(y,k)))) -> sum#(app(l,sum(cons(x,cons(y,k)))))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r9

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1) = x1
              app_A(x1,x2) = x1 + x2 + 1
              cons_A(x1,x2) = x2 + 6
              sum_A(x1) = 7
              nil_A() = 1
              plus_A(x1,x2) = x1 + x2 + 1
              zero_A() = 1
              sum#!_A(x1) = x1
              app!_A(x1,x2) = x1 + x2 + 4
              cons!_A(x1,x2) = 9
              sum!_A(x1) = 12
              nil!_A() = 8
              plus!_A(x1,x2) = x2 + 3
              zero!_A() = 7
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            app_A(x1,x2) = max{x1 + 3, x2 + 2}
            cons_A(x1,x2) = x2
            sum_A(x1) = 0
            nil_A = 0
            plus_A(x1,x2) = max{2, x2}
            zero_A = 8
            sum#!_A(x1) = x1 + 10
            app!_A(x1,x2) = max{x1 + 14, x2 + 10}
            cons!_A(x1,x2) = x2 + 2
            sum!_A(x1) = 0
            nil!_A = 1
            plus!_A(x1,x2) = max{8, x1 + 4, x2 - 32}
            zero!_A = 25
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(app) = [1, 2]
      pi(cons) = []
      pi(sum) = []
      pi(nil) = []
      pi(plus) = [1, 2]
      pi(zero) = []

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: app#(cons(x,l),k) -> app#(l,k)

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              cons_A(x1,x2) = x1 + x2 + 2
              app#!_A(x1,x2) = x1 + 7
              cons!_A(x1,x2) = x1 + 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = 0
            cons_A(x1,x2) = max{x1 + 45, x2 - 14}
            app#!_A(x1,x2) = 77
            cons!_A(x1,x2) = max{76, x1 - 13}
        
  
    partial status:
  
      pi(app#) = [2]
      pi(cons) = []

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: sum#(cons(x,cons(y,l))) -> sum#(cons(plus(x,y),l))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r1, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sum#_A(x1) = x1
              cons_A(x1,x2) = x2 + 3
              plus_A(x1,x2) = 1
              app_A(x1,x2) = x1 + x2 + 1
              nil_A() = 1
              gt_A(x1,x2) = 1
              s_A(x1) = 1
              zero_A() = 1
              true_A() = 1
              false_A() = 1
              sum#!_A(x1) = x1
              cons!_A(x1,x2) = x2 + 1
              plus!_A(x1,x2) = 7
              app!_A(x1,x2) = 5
              nil!_A() = 3
              gt!_A(x1,x2) = 5
              s!_A(x1) = 6
              zero!_A() = 1
              true!_A() = 0
              false!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sum#_A(x1) = 0
            cons_A(x1,x2) = max{45, x2 + 24}
            plus_A(x1,x2) = max{59, x1 + 24, x2 + 21}
            app_A(x1,x2) = max{70, x1 - 37, x2 + 24}
            nil_A = 56
            gt_A(x1,x2) = max{56, x1 - 1, x2 - 30}
            s_A(x1) = x1
            zero_A = 0
            true_A = 3
            false_A = 1
            sum#!_A(x1) = max{1, x1 - 35}
            cons!_A(x1,x2) = 0
            plus!_A(x1,x2) = x2 + 27
            app!_A(x1,x2) = max{93, x1 - 47, x2 + 24}
            nil!_A = 1
            gt!_A(x1,x2) = max{28, x1 - 27, x2 - 1}
            s!_A(x1) = max{26, x1 - 28}
            zero!_A = 1
            true!_A = 0
            false!_A = 0
        
  
    partial status:
  
      pi(sum#) = [1]
      pi(cons) = [2]
      pi(plus) = [1, 2]
      pi(app) = [1, 2]
      pi(nil) = []
      pi(gt) = [1]
      pi(s) = [1]
      pi(zero) = []
      pi(true) = []
      pi(false) = []

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

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r11, r12, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = 0
              s_A(x1) = 2
              if_A(x1,x2,x3) = x2 + x3 + 1
              gt_A(x1,x2) = 1
              not_A(x1) = 3
              id_A(x1) = x1 + 5
              true_A() = 1
              false_A() = 1
              zero_A() = 1
              plus#!_A(x1,x2) = 9
              s!_A(x1) = x1 + 10
              if!_A(x1,x2,x3) = 17
              gt!_A(x1,x2) = 8
              not!_A(x1) = 18
              id!_A(x1) = 14
              true!_A() = 7
              false!_A() = 7
              zero!_A() = 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{x1 + 39, x2 + 37}
            s_A(x1) = x1 + 38
            if_A(x1,x2,x3) = max{x2 + 18, x3}
            gt_A(x1,x2) = max{111, x1 - 150}
            not_A(x1) = max{37, x1 - 74}
            id_A(x1) = max{18, x1 + 13}
            true_A = 1
            false_A = 1
            zero_A = 1
            plus#!_A(x1,x2) = max{x1 + 150, x2 + 146}
            s!_A(x1) = x1 + 2
            if!_A(x1,x2,x3) = max{189, x3 - 14}
            gt!_A(x1,x2) = max{35, x1 - 2, x2 + 12}
            not!_A(x1) = max{9, x1 - 12}
            id!_A(x1) = max{0, x1 - 1}
            true!_A = 12
            false!_A = 9
            zero!_A = 13
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      pi(if) = [2, 3]
      pi(gt) = []
      pi(not) = [1]
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

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

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: 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: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x1
              id_A(x1) = x1 + 1
              s_A(x1) = x1 + 4
              if_A(x1,x2,x3) = x2 + x3 + 1
              gt_A(x1,x2) = x2 + 1
              zero_A() = 1
              true_A() = 1
              false_A() = 1
              plus#!_A(x1,x2) = 4
              id!_A(x1) = 3
              s!_A(x1) = x1 + 1
              if!_A(x1,x2,x3) = 0
              gt!_A(x1,x2) = 2
              zero!_A() = 1
              true!_A() = 0
              false!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 41}
            id_A(x1) = x1 + 40
            s_A(x1) = max{42, x1 + 26}
            if_A(x1,x2,x3) = max{27, x1 + 25, x2 + 26, x3 - 2}
            gt_A(x1,x2) = max{1, x1 - 43, x2 - 21}
            zero_A = 37
            true_A = 16
            false_A = 1
            plus#!_A(x1,x2) = max{126, x1 - 41}
            id!_A(x1) = 20
            s!_A(x1) = max{125, x1 + 84}
            if!_A(x1,x2,x3) = max{21, x1 - 33, x2 - 41, x3 - 41}
            gt!_A(x1,x2) = max{127, x1 - 41, x2 + 86}
            zero!_A = 125
            true!_A = 124
            false!_A = 87
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = [1]
      pi(s) = [1]
      pi(if) = []
      pi(gt) = []
      pi(zero) = []
      pi(true) = []
      pi(false) = []

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

and R consists of:

r1: app(nil(),k) -> k
r2: app(l,nil()) -> l
r3: app(cons(x,l),k) -> cons(x,app(l,k))
r4: sum(cons(x,nil())) -> cons(x,nil())
r5: sum(cons(x,cons(y,l))) -> sum(cons(plus(x,y),l))
r6: sum(app(l,cons(x,cons(y,k)))) -> sum(app(l,sum(cons(x,cons(y,k)))))
r7: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r8: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r9: plus(zero(),y) -> y
r10: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r11: id(x) -> x
r12: if(true(),x,y) -> x
r13: if(false(),x,y) -> y
r14: not(x) -> if(x,false(),true())
r15: gt(s(x),zero()) -> true()
r16: gt(zero(),y) -> false()
r17: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r13

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              gt#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 6
              if_A(x1,x2,x3) = x2 + x3 + 1
              false_A() = 2
              gt#!_A(x1,x2) = x1
              s!_A(x1) = 18
              if!_A(x1,x2,x3) = 14
              false!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = 0
            s_A(x1) = max{26, x1 - 27}
            if_A(x1,x2,x3) = max{49, x1 + 32, x2 - 27, x3}
            false_A = 45
            gt#!_A(x1,x2) = max{x1 - 27, x2 + 44}
            s!_A(x1) = max{48, x1 + 43}
            if!_A(x1,x2,x3) = max{47, x1 + 31, x2 - 46}
            false!_A = 0
        
  
    partial status:
  
      pi(gt#) = []
      pi(s) = []
      pi(if) = [3]
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.