Left Termination of the query pattern times_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

times(X, Y, Z) :- mult(X, Y, 0, Z).
mult(0, Y, 0, 0).
mult(s(U), Y, 0, Z) :- mult(U, Y, Y, Z).
mult(X, Y, s(W), s(Z)) :- mult(X, Y, W, Z).

Queries:

times(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:

times_in(X, Y, Z) → U1(X, Y, Z, mult_in(X, Y, 0, Z))
mult_in(X, Y, s(W), s(Z)) → U3(X, Y, W, Z, mult_in(X, Y, W, Z))
mult_in(s(U), Y, 0, Z) → U2(U, Y, Z, mult_in(U, Y, Y, Z))
mult_in(0, Y, 0, 0) → mult_out(0, Y, 0, 0)
U2(U, Y, Z, mult_out(U, Y, Y, Z)) → mult_out(s(U), Y, 0, Z)
U3(X, Y, W, Z, mult_out(X, Y, W, Z)) → mult_out(X, Y, s(W), s(Z))
U1(X, Y, Z, mult_out(X, Y, 0, Z)) → times_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
times_in(x1, x2, x3)  =  times_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
mult_in(x1, x2, x3, x4)  =  mult_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U2(x1, x2, x3, x4)  =  U2(x4)
mult_out(x1, x2, x3, x4)  =  mult_out(x4)
times_out(x1, x2, x3)  =  times_out(x3)

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:

times_in(X, Y, Z) → U1(X, Y, Z, mult_in(X, Y, 0, Z))
mult_in(X, Y, s(W), s(Z)) → U3(X, Y, W, Z, mult_in(X, Y, W, Z))
mult_in(s(U), Y, 0, Z) → U2(U, Y, Z, mult_in(U, Y, Y, Z))
mult_in(0, Y, 0, 0) → mult_out(0, Y, 0, 0)
U2(U, Y, Z, mult_out(U, Y, Y, Z)) → mult_out(s(U), Y, 0, Z)
U3(X, Y, W, Z, mult_out(X, Y, W, Z)) → mult_out(X, Y, s(W), s(Z))
U1(X, Y, Z, mult_out(X, Y, 0, Z)) → times_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
times_in(x1, x2, x3)  =  times_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
mult_in(x1, x2, x3, x4)  =  mult_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U2(x1, x2, x3, x4)  =  U2(x4)
mult_out(x1, x2, x3, x4)  =  mult_out(x4)
times_out(x1, x2, x3)  =  times_out(x3)


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

TIMES_IN(X, Y, Z) → U11(X, Y, Z, mult_in(X, Y, 0, Z))
TIMES_IN(X, Y, Z) → MULT_IN(X, Y, 0, Z)
MULT_IN(X, Y, s(W), s(Z)) → U31(X, Y, W, Z, mult_in(X, Y, W, Z))
MULT_IN(X, Y, s(W), s(Z)) → MULT_IN(X, Y, W, Z)
MULT_IN(s(U), Y, 0, Z) → U21(U, Y, Z, mult_in(U, Y, Y, Z))
MULT_IN(s(U), Y, 0, Z) → MULT_IN(U, Y, Y, Z)

The TRS R consists of the following rules:

times_in(X, Y, Z) → U1(X, Y, Z, mult_in(X, Y, 0, Z))
mult_in(X, Y, s(W), s(Z)) → U3(X, Y, W, Z, mult_in(X, Y, W, Z))
mult_in(s(U), Y, 0, Z) → U2(U, Y, Z, mult_in(U, Y, Y, Z))
mult_in(0, Y, 0, 0) → mult_out(0, Y, 0, 0)
U2(U, Y, Z, mult_out(U, Y, Y, Z)) → mult_out(s(U), Y, 0, Z)
U3(X, Y, W, Z, mult_out(X, Y, W, Z)) → mult_out(X, Y, s(W), s(Z))
U1(X, Y, Z, mult_out(X, Y, 0, Z)) → times_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
times_in(x1, x2, x3)  =  times_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
mult_in(x1, x2, x3, x4)  =  mult_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U2(x1, x2, x3, x4)  =  U2(x4)
mult_out(x1, x2, x3, x4)  =  mult_out(x4)
times_out(x1, x2, x3)  =  times_out(x3)
TIMES_IN(x1, x2, x3)  =  TIMES_IN(x1, x2)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
U21(x1, x2, x3, x4)  =  U21(x4)
MULT_IN(x1, x2, x3, x4)  =  MULT_IN(x1, x2, x3)
U11(x1, x2, x3, x4)  =  U11(x4)

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:

TIMES_IN(X, Y, Z) → U11(X, Y, Z, mult_in(X, Y, 0, Z))
TIMES_IN(X, Y, Z) → MULT_IN(X, Y, 0, Z)
MULT_IN(X, Y, s(W), s(Z)) → U31(X, Y, W, Z, mult_in(X, Y, W, Z))
MULT_IN(X, Y, s(W), s(Z)) → MULT_IN(X, Y, W, Z)
MULT_IN(s(U), Y, 0, Z) → U21(U, Y, Z, mult_in(U, Y, Y, Z))
MULT_IN(s(U), Y, 0, Z) → MULT_IN(U, Y, Y, Z)

The TRS R consists of the following rules:

times_in(X, Y, Z) → U1(X, Y, Z, mult_in(X, Y, 0, Z))
mult_in(X, Y, s(W), s(Z)) → U3(X, Y, W, Z, mult_in(X, Y, W, Z))
mult_in(s(U), Y, 0, Z) → U2(U, Y, Z, mult_in(U, Y, Y, Z))
mult_in(0, Y, 0, 0) → mult_out(0, Y, 0, 0)
U2(U, Y, Z, mult_out(U, Y, Y, Z)) → mult_out(s(U), Y, 0, Z)
U3(X, Y, W, Z, mult_out(X, Y, W, Z)) → mult_out(X, Y, s(W), s(Z))
U1(X, Y, Z, mult_out(X, Y, 0, Z)) → times_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
times_in(x1, x2, x3)  =  times_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
mult_in(x1, x2, x3, x4)  =  mult_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U2(x1, x2, x3, x4)  =  U2(x4)
mult_out(x1, x2, x3, x4)  =  mult_out(x4)
times_out(x1, x2, x3)  =  times_out(x3)
TIMES_IN(x1, x2, x3)  =  TIMES_IN(x1, x2)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
U21(x1, x2, x3, x4)  =  U21(x4)
MULT_IN(x1, x2, x3, x4)  =  MULT_IN(x1, x2, x3)
U11(x1, x2, x3, x4)  =  U11(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 4 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

MULT_IN(s(U), Y, 0, Z) → MULT_IN(U, Y, Y, Z)
MULT_IN(X, Y, s(W), s(Z)) → MULT_IN(X, Y, W, Z)

The TRS R consists of the following rules:

times_in(X, Y, Z) → U1(X, Y, Z, mult_in(X, Y, 0, Z))
mult_in(X, Y, s(W), s(Z)) → U3(X, Y, W, Z, mult_in(X, Y, W, Z))
mult_in(s(U), Y, 0, Z) → U2(U, Y, Z, mult_in(U, Y, Y, Z))
mult_in(0, Y, 0, 0) → mult_out(0, Y, 0, 0)
U2(U, Y, Z, mult_out(U, Y, Y, Z)) → mult_out(s(U), Y, 0, Z)
U3(X, Y, W, Z, mult_out(X, Y, W, Z)) → mult_out(X, Y, s(W), s(Z))
U1(X, Y, Z, mult_out(X, Y, 0, Z)) → times_out(X, Y, Z)

The argument filtering Pi contains the following mapping:
times_in(x1, x2, x3)  =  times_in(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
mult_in(x1, x2, x3, x4)  =  mult_in(x1, x2, x3)
0  =  0
s(x1)  =  s(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U2(x1, x2, x3, x4)  =  U2(x4)
mult_out(x1, x2, x3, x4)  =  mult_out(x4)
times_out(x1, x2, x3)  =  times_out(x3)
MULT_IN(x1, x2, x3, x4)  =  MULT_IN(x1, x2, x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

MULT_IN(s(U), Y, 0, Z) → MULT_IN(U, Y, Y, Z)
MULT_IN(X, Y, s(W), s(Z)) → MULT_IN(X, Y, W, Z)

R is empty.
The argument filtering Pi contains the following mapping:
0  =  0
s(x1)  =  s(x1)
MULT_IN(x1, x2, x3, x4)  =  MULT_IN(x1, x2, x3)

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
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

MULT_IN(X, Y, s(W)) → MULT_IN(X, Y, W)
MULT_IN(s(U), Y, 0) → MULT_IN(U, Y, Y)

R is empty.
Q is empty.
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: