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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1,x2) = 1
          s_A(x1) = 2
          c_A(x1) = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2) = max{x1, x2 - 1}
          s_A(x1) = x1 + 4
          c_A(x1) = max{6, x1 + 3}
          f#!_A(x1,x2) = max{5, x1 + 4, x2 + 1}
          s!_A(x1) = max{11, x1 + 8}
          c!_A(x1) = max{1, x1 - 6}
  
    partial status:
  
      pi(f#) = []
      pi(s) = []
      pi(c) = []

The following rules are removed:

  p1, p2



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1,x2) = x1 + x2 + 4
          s_A(x1) = x1 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2) = 0
          s_A(x1) = max{4, x1 - 5}
          f#!_A(x1,x2) = max{2, x1 - 6, x2 - 5}
          s!_A(x1) = max{1, x1 - 10}
  
    partial status:
  
      pi(f#) = [1, 2]
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.