Left Termination proof of /tmp/tmpgM5hei/ack.pl

Left Termination of the query pattern ack_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

ack(0, N, s(N)).
ack(s(M), 0, A) :- ack(M, s(0), A).
ack(s(M), s(N), A) :- ','(ack(s(M), N, A1), ack(M, A1, A)).

Queries:

ack(g,g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

ack_in(s(M), s(N), A) → U2(M, N, A, ack_in(s(M), N, A1))
ack_in(s(M), 0, A) → U1(M, A, ack_in(M, s(0), A))
ack_in(0, N, s(N)) → ack_out(0, N, s(N))
U1(M, A, ack_out(M, s(0), A)) → ack_out(s(M), 0, A)
U2(M, N, A, ack_out(s(M), N, A1)) → U3(M, N, A, ack_in(M, A1, A))
U3(M, N, A, ack_out(M, A1, A)) → ack_out(s(M), s(N), A)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

ack_in(s(M), s(N), A) → U2(M, N, A, ack_in(s(M), N, A1))
ack_in(s(M), 0, A) → U1(M, A, ack_in(M, s(0), A))
ack_in(0, N, s(N)) → ack_out(0, N, s(N))
U1(M, A, ack_out(M, s(0), A)) → ack_out(s(M), 0, A)
U2(M, N, A, ack_out(s(M), N, A1)) → U3(M, N, A, ack_in(M, A1, A))
U3(M, N, A, ack_out(M, A1, A)) → ack_out(s(M), s(N), A)

The argument filtering Pi contains the following mapping:
ack_in(x1, x2, x3) = ack_in(x1, x2)
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
0 = 0
U1(x1, x2, x3) = U1(x3)
ack_out(x1, x2, x3) = ack_out(x3)
U3(x1, x2, x3, x4) = U3(x4)

Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

ACK_IN(s(M), s(N), A) → U2¹(M, N, A, ack_in(s(M), N, A1))
ACK_IN(s(M), s(N), A) → ACK_IN(s(M), N, A1)
ACK_IN(s(M), 0, A) → U1¹(M, A, ack_in(M, s(0), A))
ACK_IN(s(M), 0, A) → ACK_IN(M, s(0), A)
U2¹(M, N, A, ack_out(s(M), N, A1)) → U3¹(M, N, A, ack_in(M, A1, A))
U2¹(M, N, A, ack_out(s(M), N, A1)) → ACK_IN(M, A1, A)

The TRS R consists of the following rules:

ack_in(s(M), s(N), A) → U2(M, N, A, ack_in(s(M), N, A1))
ack_in(s(M), 0, A) → U1(M, A, ack_in(M, s(0), A))
ack_in(0, N, s(N)) → ack_out(0, N, s(N))
U1(M, A, ack_out(M, s(0), A)) → ack_out(s(M), 0, A)
U2(M, N, A, ack_out(s(M), N, A1)) → U3(M, N, A, ack_in(M, A1, A))
U3(M, N, A, ack_out(M, A1, A)) → ack_out(s(M), s(N), A)

The argument filtering Pi contains the following mapping:
ack_in(x1, x2, x3) = ack_in(x1, x2)
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
0 = 0
U1(x1, x2, x3) = U1(x3)
ack_out(x1, x2, x3) = ack_out(x3)
U3(x1, x2, x3, x4) = U3(x4)
U3¹(x1, x2, x3, x4) = U3¹(x4)
ACK_IN(x1, x2, x3) = ACK_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x1, x4)
U1¹(x1, x2, x3) = U1¹(x3)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

ACK_IN(s(M), s(N), A) → U2¹(M, N, A, ack_in(s(M), N, A1))
ACK_IN(s(M), s(N), A) → ACK_IN(s(M), N, A1)
ACK_IN(s(M), 0, A) → U1¹(M, A, ack_in(M, s(0), A))
ACK_IN(s(M), 0, A) → ACK_IN(M, s(0), A)
U2¹(M, N, A, ack_out(s(M), N, A1)) → U3¹(M, N, A, ack_in(M, A1, A))
U2¹(M, N, A, ack_out(s(M), N, A1)) → ACK_IN(M, A1, A)

The TRS R consists of the following rules:

ack_in(s(M), s(N), A) → U2(M, N, A, ack_in(s(M), N, A1))
ack_in(s(M), 0, A) → U1(M, A, ack_in(M, s(0), A))
ack_in(0, N, s(N)) → ack_out(0, N, s(N))
U1(M, A, ack_out(M, s(0), A)) → ack_out(s(M), 0, A)
U2(M, N, A, ack_out(s(M), N, A1)) → U3(M, N, A, ack_in(M, A1, A))
U3(M, N, A, ack_out(M, A1, A)) → ack_out(s(M), s(N), A)

The argument filtering Pi contains the following mapping:
ack_in(x1, x2, x3) = ack_in(x1, x2)
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
0 = 0
U1(x1, x2, x3) = U1(x3)
ack_out(x1, x2, x3) = ack_out(x3)
U3(x1, x2, x3, x4) = U3(x4)
U3¹(x1, x2, x3, x4) = U3¹(x4)
ACK_IN(x1, x2, x3) = ACK_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x1, x4)
U1¹(x1, x2, x3) = U1¹(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

ACK_IN(s(M), s(N), A) → U2¹(M, N, A, ack_in(s(M), N, A1))
ACK_IN(s(M), 0, A) → ACK_IN(M, s(0), A)
ACK_IN(s(M), s(N), A) → ACK_IN(s(M), N, A1)
U2¹(M, N, A, ack_out(s(M), N, A1)) → ACK_IN(M, A1, A)

The TRS R consists of the following rules:

ack_in(s(M), s(N), A) → U2(M, N, A, ack_in(s(M), N, A1))
ack_in(s(M), 0, A) → U1(M, A, ack_in(M, s(0), A))
ack_in(0, N, s(N)) → ack_out(0, N, s(N))
U1(M, A, ack_out(M, s(0), A)) → ack_out(s(M), 0, A)
U2(M, N, A, ack_out(s(M), N, A1)) → U3(M, N, A, ack_in(M, A1, A))
U3(M, N, A, ack_out(M, A1, A)) → ack_out(s(M), s(N), A)

The argument filtering Pi contains the following mapping:
ack_in(x1, x2, x3) = ack_in(x1, x2)
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x1, x4)
0 = 0
U1(x1, x2, x3) = U1(x3)
ack_out(x1, x2, x3) = ack_out(x3)
U3(x1, x2, x3, x4) = U3(x4)
ACK_IN(x1, x2, x3) = ACK_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x1, x4)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

U2¹(M, ack_out(A1)) → ACK_IN(M, A1)
ACK_IN(s(M), s(N)) → ACK_IN(s(M), N)
ACK_IN(s(M), 0) → ACK_IN(M, s(0))
ACK_IN(s(M), s(N)) → U2¹(M, ack_in(s(M), N))

The TRS R consists of the following rules:

ack_in(s(M), s(N)) → U2(M, ack_in(s(M), N))
ack_in(s(M), 0) → U1(ack_in(M, s(0)))
ack_in(0, N) → ack_out(s(N))
U1(ack_out(A)) → ack_out(A)
U2(M, ack_out(A1)) → U3(ack_in(M, A1))
U3(ack_out(A)) → ack_out(A)

The set Q consists of the following terms:

ack_in(x0, x1)
U1(x0)
U2(x0, x1)
U3(x0)

We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

U2¹(M, ack_out(A1)) → ACK_IN(M, A1)
The graph contains the following edges 1 >= 1, 2 > 2
ACK_IN(s(M), s(N)) → U2¹(M, ack_in(s(M), N))
The graph contains the following edges 1 > 1
ACK_IN(s(M), s(N)) → ACK_IN(s(M), N)
The graph contains the following edges 1 >= 1, 2 > 2
ACK_IN(s(M), 0) → ACK_IN(M, s(0))
The graph contains the following edges 1 > 1