Left Termination of the query pattern append_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

append(.(H, X), Y, .(X, Z)) :- append(X, Y, Z).
append([], Y, Y).

Queries:

append(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:

append_in([], Y, Y) → append_out([], Y, Y)
append_in(.(H, X), Y, .(X, Z)) → U1(H, X, Y, Z, append_in(X, Y, Z))
U1(H, X, Y, Z, append_out(X, Y, Z)) → append_out(.(H, X), Y, .(X, Z))

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

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:

append_in([], Y, Y) → append_out([], Y, Y)
append_in(.(H, X), Y, .(X, Z)) → U1(H, X, Y, Z, append_in(X, Y, Z))
U1(H, X, Y, Z, append_out(X, Y, Z)) → append_out(.(H, X), Y, .(X, Z))

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


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

APPEND_IN(.(H, X), Y, .(X, Z)) → U11(H, X, Y, Z, append_in(X, Y, Z))
APPEND_IN(.(H, X), Y, .(X, Z)) → APPEND_IN(X, Y, Z)

The TRS R consists of the following rules:

append_in([], Y, Y) → append_out([], Y, Y)
append_in(.(H, X), Y, .(X, Z)) → U1(H, X, Y, Z, append_in(X, Y, Z))
U1(H, X, Y, Z, append_out(X, Y, Z)) → append_out(.(H, X), Y, .(X, Z))

The argument filtering Pi contains the following mapping:
append_in(x1, x2, x3)  =  append_in(x1, x2)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x2, x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN(x1, x2)
U11(x1, x2, x3, x4, x5)  =  U11(x2, 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:

APPEND_IN(.(H, X), Y, .(X, Z)) → U11(H, X, Y, Z, append_in(X, Y, Z))
APPEND_IN(.(H, X), Y, .(X, Z)) → APPEND_IN(X, Y, Z)

The TRS R consists of the following rules:

append_in([], Y, Y) → append_out([], Y, Y)
append_in(.(H, X), Y, .(X, Z)) → U1(H, X, Y, Z, append_in(X, Y, Z))
U1(H, X, Y, Z, append_out(X, Y, Z)) → append_out(.(H, X), Y, .(X, Z))

The argument filtering Pi contains the following mapping:
append_in(x1, x2, x3)  =  append_in(x1, x2)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x2, x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN(x1, x2)
U11(x1, x2, x3, x4, x5)  =  U11(x2, 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:

APPEND_IN(.(H, X), Y, .(X, Z)) → APPEND_IN(X, Y, Z)

The TRS R consists of the following rules:

append_in([], Y, Y) → append_out([], Y, Y)
append_in(.(H, X), Y, .(X, Z)) → U1(H, X, Y, Z, append_in(X, Y, Z))
U1(H, X, Y, Z, append_out(X, Y, Z)) → append_out(.(H, X), Y, .(X, Z))

The argument filtering Pi contains the following mapping:
append_in(x1, x2, x3)  =  append_in(x1, x2)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x2, x5)
APPEND_IN(x1, x2, x3)  =  APPEND_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:

APPEND_IN(.(H, X), Y, .(X, Z)) → APPEND_IN(X, Y, Z)

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

APPEND_IN(.(H, X), Y) → APPEND_IN(X, 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: