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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      h#_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,1),(0,1)) x2
      c_A(x1,x2) = x1 + x2 + (1,3)
      s_A(x1) = ((0,1),(0,0)) x1 + (1,1)
      |0|_A() = (1,4)

The next rules are strictly ordered:

  p1, p2

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