NO

Problem:
 f(a(x),x) -> f(x,a(x))
 f(b(x),x) -> f(x,b(x))
 g(b(x),y) -> g(a(a(x)),y)
 g(c(x),y) -> y
 a(x) -> b(x)

Proof:
 Matrix Interpretation Processor:
  dimension: 1
  interpretation:
   [c](x0) = 4x0 + 3,
   
   [g](x0, x1) = x0 + x1 + 2,
   
   [b](x0) = x0,
   
   [f](x0, x1) = x0 + x1,
   
   [a](x0) = x0
  orientation:
   f(a(x),x) = 2x >= 2x = f(x,a(x))
   
   f(b(x),x) = 2x >= 2x = f(x,b(x))
   
   g(b(x),y) = x + y + 2 >= x + y + 2 = g(a(a(x)),y)
   
   g(c(x),y) = 4x + y + 5 >= y = y
   
   a(x) = x >= x = b(x)
  problem:
   f(a(x),x) -> f(x,a(x))
   f(b(x),x) -> f(x,b(x))
   g(b(x),y) -> g(a(a(x)),y)
   a(x) -> b(x)
  Unfolding Processor:
   loop length: 2
   terms:
    g(b(x),y)
    g(a(a(x)),y)
   context: []
   substitution:
    x -> a(x)
    y -> y
   Qed