YES

We show the termination of the TRS R:

  f(c(X,s(Y))) -> f(c(s(X),Y))
  g(c(s(X),Y)) -> f(c(X,s(Y)))

-- SCC decomposition.

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

p1: f#(c(X,s(Y))) -> f#(c(s(X),Y))
p2: g#(c(s(X),Y)) -> f#(c(X,s(Y)))

and R consists of:

r1: f(c(X,s(Y))) -> f(c(s(X),Y))
r2: g(c(s(X),Y)) -> f(c(X,s(Y)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(c(X,s(Y))) -> f#(c(s(X),Y))

and R consists of:

r1: f(c(X,s(Y))) -> f(c(s(X),Y))
r2: g(c(s(X),Y)) -> f(c(X,s(Y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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