YES

TRS:

  ap(ap(ff(),x),x) -> ap(ap(x,ap(ff(),x)),ap(ap(cons(),x),nil()))

max/plus interpretations on N:

  ap_A(x1,x2) = max{3, -16 + x1, -5 + x2}
  ap#_A(x1,x2) = max{0, 1 + x1, 1 + x2}
  ff_A = 23
  ff#_A = 3
  cons_A = 7
  cons#_A = 2
  nil_A = 2
  nil#_A = 7

precedence:

  ap > nil > ff > cons