YES

We show the termination of the TRS R:

  h(x,c(y,z)) -> h(c(s(y),x),z)
  h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

-- SCC decomposition.

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

p1: h#(x,c(y,z)) -> h#(c(s(y),x),z)
p2: h#(c(s(x),c(s(|0|()),y)),z) -> h#(y,c(s(|0|()),c(x,z)))

and R consists of:

r1: h(x,c(y,z)) -> h(c(s(y),x),z)
r2: h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: h#(x,c(y,z)) -> h#(c(s(y),x),z)
p2: h#(c(s(x),c(s(|0|()),y)),z) -> h#(y,c(s(|0|()),c(x,z)))

and R consists of:

r1: h(x,c(y,z)) -> h(c(s(y),x),z)
r2: h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          h#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,32)
          c_A(x1,x2) = ((0,1),(1,0)) x1 + x2 + (13,10)
          s_A(x1) = ((0,1),(1,0)) x1 + (1,2)
          |0|_A() = (2,9)
  
    precedence:
    
      h# = c = s = |0|
    
    partial status:
  
      pi(h#) = []
      pi(c) = [2]
      pi(s) = []
      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: h#(x,c(y,z)) -> h#(c(s(y),x),z)

and R consists of:

r1: h(x,c(y,z)) -> h(c(s(y),x),z)
r2: h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: h#(x,c(y,z)) -> h#(c(s(y),x),z)

and R consists of:

r1: h(x,c(y,z)) -> h(c(s(y),x),z)
r2: h(c(s(x),c(s(|0|()),y)),z) -> h(y,c(s(|0|()),c(x,z)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          h#_A(x1,x2) = ((0,1),(0,0)) x2 + (2,2)
          c_A(x1,x2) = ((0,0),(1,1)) x2 + (1,1)
          s_A(x1) = (2,2)
  
    precedence:
    
      h# = c = s
    
    partial status:
  
      pi(h#) = []
      pi(c) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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