YES

We show the termination of the TRS R:

  +(x,|0|()) -> x
  +(x,s(y)) -> s(+(x,y))
  +(|0|(),y) -> y
  +(s(x),y) -> s(+(x,y))
  +(x,+(y,z)) -> +(+(x,y),z)
  f(g(f(x))) -> f(h(s(|0|()),x))
  f(g(h(x,y))) -> f(h(s(x),y))
  f(h(x,h(y,z))) -> f(h(+(x,y),z))

-- SCC decomposition.

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

p1: +#(x,s(y)) -> +#(x,y)
p2: +#(s(x),y) -> +#(x,y)
p3: +#(x,+(y,z)) -> +#(+(x,y),z)
p4: +#(x,+(y,z)) -> +#(x,y)
p5: f#(g(f(x))) -> f#(h(s(|0|()),x))
p6: f#(g(h(x,y))) -> f#(h(s(x),y))
p7: f#(h(x,h(y,z))) -> f#(h(+(x,y),z))
p8: f#(h(x,h(y,z))) -> +#(x,y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,s(y)) -> s(+(x,y))
r3: +(|0|(),y) -> y
r4: +(s(x),y) -> s(+(x,y))
r5: +(x,+(y,z)) -> +(+(x,y),z)
r6: f(g(f(x))) -> f(h(s(|0|()),x))
r7: f(g(h(x,y))) -> f(h(s(x),y))
r8: f(h(x,h(y,z))) -> f(h(+(x,y),z))

The estimated dependency graph contains the following SCCs:

  {p7}
  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: f#(h(x,h(y,z))) -> f#(h(+(x,y),z))

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,s(y)) -> s(+(x,y))
r3: +(|0|(),y) -> y
r4: +(s(x),y) -> s(+(x,y))
r5: +(x,+(y,z)) -> +(+(x,y),z)
r6: f(g(f(x))) -> f(h(s(|0|()),x))
r7: f(g(h(x,y))) -> f(h(s(x),y))
r8: f(h(x,h(y,z))) -> f(h(+(x,y),z))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1) = x1
        h_A(x1,x2) = x2 + 1
        +_A(x1,x2) = x1 + x2 + 1
        |0|_A() = 0
        s_A(x1) = 0
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > |0| > + > f# > h
      
      argument filter:
    
        pi(f#) = 1
        pi(h) = [2]
        pi(+) = []
        pi(|0|) = []
        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: +#(x,s(y)) -> +#(x,y)
p2: +#(x,+(y,z)) -> +#(x,y)
p3: +#(x,+(y,z)) -> +#(+(x,y),z)
p4: +#(s(x),y) -> +#(x,y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,s(y)) -> s(+(x,y))
r3: +(|0|(),y) -> y
r4: +(s(x),y) -> s(+(x,y))
r5: +(x,+(y,z)) -> +(+(x,y),z)
r6: f(g(f(x))) -> f(h(s(|0|()),x))
r7: f(g(h(x,y))) -> f(h(s(x),y))
r8: f(h(x,h(y,z))) -> f(h(+(x,y),z))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
        +_A(x1,x2) = x1 + x2 + 1
        |0|_A() = 0
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        + > +# > |0| > s
      
      argument filter:
    
        pi(+#) = [2]
        pi(s) = 1
        pi(+) = [2]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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