Left Termination of the query pattern interleave_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

interleave([], Xs, Xs).
interleave(.(X, Xs), Ys, .(X, Zs)) :- interleave(Ys, Xs, Zs).

Queries:

interleave(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:

interleave_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, interleave_in(Ys, Xs, Zs))
interleave_in([], Xs, Xs) → interleave_out([], Xs, Xs)
U1(X, Xs, Ys, Zs, interleave_out(Ys, Xs, Zs)) → interleave_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_in(x1, x2, x3)  =  interleave_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
interleave_out(x1, x2, x3)  =  interleave_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:

interleave_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, interleave_in(Ys, Xs, Zs))
interleave_in([], Xs, Xs) → interleave_out([], Xs, Xs)
U1(X, Xs, Ys, Zs, interleave_out(Ys, Xs, Zs)) → interleave_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_in(x1, x2, x3)  =  interleave_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
interleave_out(x1, x2, x3)  =  interleave_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:

INTERLEAVE_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, interleave_in(Ys, Xs, Zs))
INTERLEAVE_IN(.(X, Xs), Ys, .(X, Zs)) → INTERLEAVE_IN(Ys, Xs, Zs)

The TRS R consists of the following rules:

interleave_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, interleave_in(Ys, Xs, Zs))
interleave_in([], Xs, Xs) → interleave_out([], Xs, Xs)
U1(X, Xs, Ys, Zs, interleave_out(Ys, Xs, Zs)) → interleave_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_in(x1, x2, x3)  =  interleave_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
interleave_out(x1, x2, x3)  =  interleave_out(x3)
INTERLEAVE_IN(x1, x2, x3)  =  INTERLEAVE_IN(x1, x2)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

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:

INTERLEAVE_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, interleave_in(Ys, Xs, Zs))
INTERLEAVE_IN(.(X, Xs), Ys, .(X, Zs)) → INTERLEAVE_IN(Ys, Xs, Zs)

The TRS R consists of the following rules:

interleave_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, interleave_in(Ys, Xs, Zs))
interleave_in([], Xs, Xs) → interleave_out([], Xs, Xs)
U1(X, Xs, Ys, Zs, interleave_out(Ys, Xs, Zs)) → interleave_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_in(x1, x2, x3)  =  interleave_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
interleave_out(x1, x2, x3)  =  interleave_out(x3)
INTERLEAVE_IN(x1, x2, x3)  =  INTERLEAVE_IN(x1, x2)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

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

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

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

INTERLEAVE_IN(.(X, Xs), Ys, .(X, Zs)) → INTERLEAVE_IN(Ys, Xs, Zs)

The TRS R consists of the following rules:

interleave_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, interleave_in(Ys, Xs, Zs))
interleave_in([], Xs, Xs) → interleave_out([], Xs, Xs)
U1(X, Xs, Ys, Zs, interleave_out(Ys, Xs, Zs)) → interleave_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
interleave_in(x1, x2, x3)  =  interleave_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
interleave_out(x1, x2, x3)  =  interleave_out(x3)
INTERLEAVE_IN(x1, x2, x3)  =  INTERLEAVE_IN(x1, x2)

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:

INTERLEAVE_IN(.(X, Xs), Ys, .(X, Zs)) → INTERLEAVE_IN(Ys, Xs, Zs)

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

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:

INTERLEAVE_IN(.(X, Xs), Ys) → INTERLEAVE_IN(Ys, Xs)

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: