YES

TRS:

  b(b(0(),y),x) -> y
  c(c(c(y))) -> c(c(a(a(c(b(0(),y)),0()),0())))
  a(y,0()) -> b(y,0())

max/plus interpretations on N:

  b_A(x1,x2) = max{26, -16 + x1, 16 + x2}
  b#_A(x1,x2) = max{0, 0, 0}
  0_A = 10
  0#_A = 4
  c_A(x1) = max{27, 21 + x1}
  c#_A(x1) = max{2, 6 + x1}
  a_A(x1,x2) = max{26, -16 + x1, -26}
  a#_A(x1,x2) = max{5, 1, 3}

precedence:

  c > a > b = 0