Left Termination of the query pattern append3_in_3(a, a, g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

append1([], X, X).
append1(.(X, Y), U, .(X, Z)) :- append1(Y, U, Z).
append2([], X, X).
append2(.(X, Y), U, .(X, Z)) :- append2(Y, U, Z).
append3([], X, X).
append3(.(X, Y), U, .(X, Z)) :- append3(Y, U, Z).

Queries:

append3(a,a,g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
append3_in: (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

append3_in_aag([], X, X) → append3_out_aag([], X, X)
append3_in_aag(.(X, Y), U, .(X, Z)) → U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) → append3_out_aag(.(X, Y), U, .(X, Z))

The argument filtering Pi contains the following mapping:
append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, 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:

append3_in_aag([], X, X) → append3_out_aag([], X, X)
append3_in_aag(.(X, Y), U, .(X, Z)) → U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) → append3_out_aag(.(X, Y), U, .(X, Z))

The argument filtering Pi contains the following mapping:
append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, 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:

APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) → U3_AAG(X, Y, U, Z, append3_in_aag(Y, U, Z))
APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) → APPEND3_IN_AAG(Y, U, Z)

The TRS R consists of the following rules:

append3_in_aag([], X, X) → append3_out_aag([], X, X)
append3_in_aag(.(X, Y), U, .(X, Z)) → U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) → append3_out_aag(.(X, Y), U, .(X, Z))

The argument filtering Pi contains the following mapping:
append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
APPEND3_IN_AAG(x1, x2, x3)  =  APPEND3_IN_AAG(x3)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(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:

APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) → U3_AAG(X, Y, U, Z, append3_in_aag(Y, U, Z))
APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) → APPEND3_IN_AAG(Y, U, Z)

The TRS R consists of the following rules:

append3_in_aag([], X, X) → append3_out_aag([], X, X)
append3_in_aag(.(X, Y), U, .(X, Z)) → U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) → append3_out_aag(.(X, Y), U, .(X, Z))

The argument filtering Pi contains the following mapping:
append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
APPEND3_IN_AAG(x1, x2, x3)  =  APPEND3_IN_AAG(x3)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(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:

APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) → APPEND3_IN_AAG(Y, U, Z)

The TRS R consists of the following rules:

append3_in_aag([], X, X) → append3_out_aag([], X, X)
append3_in_aag(.(X, Y), U, .(X, Z)) → U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) → append3_out_aag(.(X, Y), U, .(X, Z))

The argument filtering Pi contains the following mapping:
append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
APPEND3_IN_AAG(x1, x2, x3)  =  APPEND3_IN_AAG(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:

APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) → APPEND3_IN_AAG(Y, U, Z)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND3_IN_AAG(x1, x2, x3)  =  APPEND3_IN_AAG(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:

APPEND3_IN_AAG(.(X, Z)) → APPEND3_IN_AAG(Z)

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: