YES

We show the termination of the TRS R:

  f(s(x),y) -> f(x,s(x))
  f(x,s(y)) -> f(y,x)
  f(c(x),y) -> f(x,s(x))

-- SCC decomposition.

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

p1: f#(s(x),y) -> f#(x,s(x))
p2: f#(x,s(y)) -> f#(y,x)
p3: f#(c(x),y) -> f#(x,s(x))

and R consists of:

r1: f(s(x),y) -> f(x,s(x))
r2: f(x,s(y)) -> f(y,x)
r3: f(c(x),y) -> f(x,s(x))

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: f#(s(x),y) -> f#(x,s(x))
p2: f#(c(x),y) -> f#(x,s(x))
p3: f#(x,s(y)) -> f#(y,x)

and R consists of:

r1: f(s(x),y) -> f(x,s(x))
r2: f(x,s(y)) -> f(y,x)
r3: f(c(x),y) -> f(x,s(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(1,1)) x1 + x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        c_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        c_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3

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