YES

We show the termination of the TRS R:

  a(h(),h(),h(),x) -> s(x)
  a(l,x,s(y),h()) -> a(l,x,y,s(h()))
  a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
  a(l,s(x),h(),z) -> a(l,x,z,z)
  a(s(l),h(),h(),z) -> a(l,z,h(),z)
  +(x,h()) -> x
  +(h(),x) -> x
  +(s(x),s(y)) -> s(s(+(x,y)))
  +(+(x,y),z) -> +(x,+(y,z))
  s(h()) -> |1|()
  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(a(x,y,h(),h()),l))

-- SCC decomposition.

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

p1: a#(h(),h(),h(),x) -> s#(x)
p2: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p3: a#(l,x,s(y),h()) -> s#(h())
p4: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))
p5: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p6: a#(l,s(x),h(),z) -> a#(l,x,z,z)
p7: a#(s(l),h(),h(),z) -> a#(l,z,h(),z)
p8: +#(s(x),s(y)) -> s#(s(+(x,y)))
p9: +#(s(x),s(y)) -> s#(+(x,y))
p10: +#(s(x),s(y)) -> +#(x,y)
p11: +#(+(x,y),z) -> +#(x,+(y,z))
p12: +#(+(x,y),z) -> +#(y,z)
p13: app#(cons(x,l),k) -> app#(l,k)
p14: sum#(cons(x,cons(y,l))) -> sum#(cons(a(x,y,h(),h()),l))
p15: sum#(cons(x,cons(y,l))) -> a#(x,y,h(),h())

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The estimated dependency graph contains the following SCCs:

  {p14}
  {p2, p4, p5, p6, p7}
  {p10, p11, p12}
  {p13}


-- Reduction pair.

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

p1: sum#(cons(x,cons(y,l))) -> sum#(cons(a(x,y,h(),h()),l))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        sum#_A(x1) = max{2, x1 - 5}
        cons_A(x1,x2) = max{6, x2 + 2}
        a_A(x1,x2,x3,x4) = max{x1 + 1, x2, x3, x4}
        h_A = 0
        s_A(x1) = x1
        |1|_A = 0
  
    precedence:
    
      sum# = cons = a > s = |1| > h
    
    partial status:
  
      pi(sum#) = []
      pi(cons) = [2]
      pi(a) = [1, 2, 3, 4]
      pi(h) = []
      pi(s) = [1]
      pi(|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: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(s(l),h(),h(),z) -> a#(l,z,h(),z)
p3: a#(l,s(x),h(),z) -> a#(l,x,z,z)
p4: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p5: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3,x4) = x1 + 3
        s_A(x1) = x1
        h_A = 0
        a_A(x1,x2,x3,x4) = max{x1 + 4, x2, x3, x4}
        |1|_A = 0
  
    precedence:
    
      a > a# = s = |1| > h
    
    partial status:
  
      pi(a#) = [1]
      pi(s) = [1]
      pi(h) = []
      pi(a) = [1, 2, 3, 4]
      pi(|1|) = []

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: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,s(x),h(),z) -> a#(l,x,z,z)
p3: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p4: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))
p3: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p4: a#(l,s(x),h(),z) -> a#(l,x,z,z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3,x4) = max{x1 + 10, x2 + 7, x3 + 6, x4 + 6}
        s_A(x1) = max{4, x1}
        h_A = 2
        a_A(x1,x2,x3,x4) = max{x1 + 4, x2, x3, x4}
        |1|_A = 3
  
    precedence:
    
      a# > a > h > s = |1|
    
    partial status:
  
      pi(a#) = [2]
      pi(s) = [1]
      pi(h) = []
      pi(a) = [1, 2, 3, 4]
      pi(|1|) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))
p3: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)
p3: a#(l,x,s(y),s(z)) -> a#(l,x,y,a(l,x,s(y),z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3,x4) = max{x1 + 8, x2 + 7, x3 + 8, x4 + 4}
        s_A(x1) = max{3, x1}
        h_A = 0
        a_A(x1,x2,x3,x4) = max{x1 + 4, x2, x3, x4}
        |1|_A = 1
  
    precedence:
    
      a# > a > s = |1| > h
    
    partial status:
  
      pi(a#) = [3]
      pi(s) = [1]
      pi(h) = []
      pi(a) = [1, 2, 3, 4]
      pi(|1|) = []

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: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: a#(l,x,s(y),h()) -> a#(l,x,y,s(h()))
p2: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3,x4) = max{10, x1 + 7, x3 + 7, x4 + 4}
        s_A(x1) = x1 + 3
        h_A = 0
        |1|_A = 0
  
    precedence:
    
      s = h > a# > |1|
    
    partial status:
  
      pi(a#) = [3]
      pi(s) = [1]
      pi(h) = []
      pi(|1|) = []

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: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: a#(l,x,s(y),s(z)) -> a#(l,x,s(y),z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2,x3,x4) = max{0, x4 - 3}
        s_A(x1) = max{4, x1 + 2}
        h_A = 4
        |1|_A = 5
  
    precedence:
    
      a# = s = h = |1|
    
    partial status:
  
      pi(a#) = []
      pi(s) = [1]
      pi(h) = []
      pi(|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: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)
p3: +#(+(x,y),z) -> +#(x,+(y,z))

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  r6, r7, r8, r9, r10

Take the reduction pair:

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

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: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: +#(s(x),s(y)) -> +#(x,y)
p2: +#(+(x,y),z) -> +#(y,z)

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

and R consists of:

r1: a(h(),h(),h(),x) -> s(x)
r2: a(l,x,s(y),h()) -> a(l,x,y,s(h()))
r3: a(l,x,s(y),s(z)) -> a(l,x,y,a(l,x,s(y),z))
r4: a(l,s(x),h(),z) -> a(l,x,z,z)
r5: a(s(l),h(),h(),z) -> a(l,z,h(),z)
r6: +(x,h()) -> x
r7: +(h(),x) -> x
r8: +(s(x),s(y)) -> s(s(+(x,y)))
r9: +(+(x,y),z) -> +(x,+(y,z))
r10: s(h()) -> |1|()
r11: app(nil(),k) -> k
r12: app(l,nil()) -> l
r13: app(cons(x,l),k) -> cons(x,app(l,k))
r14: sum(cons(x,nil())) -> cons(x,nil())
r15: sum(cons(x,cons(y,l))) -> sum(cons(a(x,y,h(),h()),l))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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