The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the following initial DP problem:

  P1. (1) plus#(s(Y), U) => plus#(Y, s(U))
      (2) plus#(plus(V, W), P) => plus#(W, P)
      (3) plus#(plus(V, W), P) => plus#(V, plus(W, P))
      (4) mult#(s(Y1), U1) => mult#(Y1, U1)
      (5) mult#(s(Y1), U1) => plus#(mult(Y1, U1), U1)
      (6) mult#(plus(V1, W1), P1) => mult#(V1, P1)
      (7) mult#(plus(V1, W1), P1) => mult#(W1, P1)
      (8) mult#(plus(V1, W1), P1) => plus#(mult(V1, P1), mult(W1, P1))

***** We apply the Graph Processor on P1.

Considering the 2 SCCs, this DP problem is split into the following new problems.

  P2. (1) plus#(s(Y), U) => plus#(Y, s(U))
      (2) plus#(plus(V, W), P) => plus#(W, P)
      (3) plus#(plus(V, W), P) => plus#(V, plus(W, P))

  P3. (1) mult#(s(Y1), U1) => mult#(Y1, U1)
      (2) mult#(plus(V1, W1), P1) => mult#(V1, P1)
      (3) mult#(plus(V1, W1), P1) => mult#(W1, P1)

***** We apply the Subterm Criterion Processor on P2.

We use the following projection function:

  nu(plus#) = 1

We thus have:

  (1) s(Y)       |>| Y
  (2) plus(V, W) |>| W
  (3) plus(V, W) |>| V

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

***** We apply the Subterm Criterion Processor on P3.

We use the following projection function:

  nu(mult#) = 1

We thus have:

  (1) s(Y1)        |>| Y1
  (2) plus(V1, W1) |>| V1
  (3) plus(V1, W1) |>| W1

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.