Left Termination proof of /tmp/tmphb5Qr8/sameleaves.pl

Left Termination of the query pattern sameleaves_in_2(g, g) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

sameleaves(leaf(L), leaf(L)).
sameleaves(tree(T1, T2), tree(S1, S2)) :- ','(getleave(T1, T2, L, T), ','(getleave(S1, S2, L, S), sameleaves(T, S))).
getleave(leaf(A), C, A, C).
getleave(tree(A, B), C, L, O) :- getleave(A, tree(B, C), L, O).

Queries:

sameleaves(g,g).

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:

sameleaves_in(tree(T1, T2), tree(S1, S2)) → U1(T1, T2, S1, S2, getleave_in(T1, T2, L, T))
getleave_in(tree(A, B), C, L, O) → U4(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
getleave_in(leaf(A), C, A, C) → getleave_out(leaf(A), C, A, C)
U4(A, B, C, L, O, getleave_out(A, tree(B, C), L, O)) → getleave_out(tree(A, B), C, L, O)
U1(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U2(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → U3(T1, T2, S1, S2, sameleaves_in(T, S))
sameleaves_in(leaf(L), leaf(L)) → sameleaves_out(leaf(L), leaf(L))
U3(T1, T2, S1, S2, sameleaves_out(T, S)) → sameleaves_out(tree(T1, T2), tree(S1, S2))

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:

sameleaves_in(tree(T1, T2), tree(S1, S2)) → U1(T1, T2, S1, S2, getleave_in(T1, T2, L, T))
getleave_in(tree(A, B), C, L, O) → U4(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
getleave_in(leaf(A), C, A, C) → getleave_out(leaf(A), C, A, C)
U4(A, B, C, L, O, getleave_out(A, tree(B, C), L, O)) → getleave_out(tree(A, B), C, L, O)
U1(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U2(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → U3(T1, T2, S1, S2, sameleaves_in(T, S))
sameleaves_in(leaf(L), leaf(L)) → sameleaves_out(leaf(L), leaf(L))
U3(T1, T2, S1, S2, sameleaves_out(T, S)) → sameleaves_out(tree(T1, T2), tree(S1, S2))

The argument filtering Pi contains the following mapping:
sameleaves_in(x1, x2) = sameleaves_in(x1, x2)
tree(x1, x2) = tree(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x4, x5)
getleave_in(x1, x2, x3, x4) = getleave_in(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
leaf(x1) = leaf(x1)
getleave_out(x1, x2, x3, x4) = getleave_out(x3, x4)
U2(x1, x2, x3, x4, x5, x6) = U2(x5, x6)
U3(x1, x2, x3, x4, x5) = U3(x5)
sameleaves_out(x1, x2) = sameleaves_out

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

SAMELEAVES_IN(tree(T1, T2), tree(S1, S2)) → U1¹(T1, T2, S1, S2, getleave_in(T1, T2, L, T))
SAMELEAVES_IN(tree(T1, T2), tree(S1, S2)) → GETLEAVE_IN(T1, T2, L, T)
GETLEAVE_IN(tree(A, B), C, L, O) → U4¹(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
GETLEAVE_IN(tree(A, B), C, L, O) → GETLEAVE_IN(A, tree(B, C), L, O)
U1¹(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2¹(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U1¹(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → GETLEAVE_IN(S1, S2, L, S)
U2¹(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → U3¹(T1, T2, S1, S2, sameleaves_in(T, S))
U2¹(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → SAMELEAVES_IN(T, S)

The TRS R consists of the following rules:

sameleaves_in(tree(T1, T2), tree(S1, S2)) → U1(T1, T2, S1, S2, getleave_in(T1, T2, L, T))
getleave_in(tree(A, B), C, L, O) → U4(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
getleave_in(leaf(A), C, A, C) → getleave_out(leaf(A), C, A, C)
U4(A, B, C, L, O, getleave_out(A, tree(B, C), L, O)) → getleave_out(tree(A, B), C, L, O)
U1(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U2(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → U3(T1, T2, S1, S2, sameleaves_in(T, S))
sameleaves_in(leaf(L), leaf(L)) → sameleaves_out(leaf(L), leaf(L))
U3(T1, T2, S1, S2, sameleaves_out(T, S)) → sameleaves_out(tree(T1, T2), tree(S1, S2))

The argument filtering Pi contains the following mapping:
sameleaves_in(x1, x2) = sameleaves_in(x1, x2)
tree(x1, x2) = tree(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x4, x5)
getleave_in(x1, x2, x3, x4) = getleave_in(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
leaf(x1) = leaf(x1)
getleave_out(x1, x2, x3, x4) = getleave_out(x3, x4)
U2(x1, x2, x3, x4, x5, x6) = U2(x5, x6)
U3(x1, x2, x3, x4, x5) = U3(x5)
sameleaves_out(x1, x2) = sameleaves_out
U3¹(x1, x2, x3, x4, x5) = U3¹(x5)
U2¹(x1, x2, x3, x4, x5, x6) = U2¹(x5, x6)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
SAMELEAVES_IN(x1, x2) = SAMELEAVES_IN(x1, x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x4, x5)
GETLEAVE_IN(x1, x2, x3, x4) = GETLEAVE_IN(x1, x2)

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:

SAMELEAVES_IN(tree(T1, T2), tree(S1, S2)) → U1¹(T1, T2, S1, S2, getleave_in(T1, T2, L, T))
SAMELEAVES_IN(tree(T1, T2), tree(S1, S2)) → GETLEAVE_IN(T1, T2, L, T)
GETLEAVE_IN(tree(A, B), C, L, O) → U4¹(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
GETLEAVE_IN(tree(A, B), C, L, O) → GETLEAVE_IN(A, tree(B, C), L, O)
U1¹(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2¹(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U1¹(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → GETLEAVE_IN(S1, S2, L, S)
U2¹(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → U3¹(T1, T2, S1, S2, sameleaves_in(T, S))
U2¹(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → SAMELEAVES_IN(T, S)

The TRS R consists of the following rules:

sameleaves_in(tree(T1, T2), tree(S1, S2)) → U1(T1, T2, S1, S2, getleave_in(T1, T2, L, T))
getleave_in(tree(A, B), C, L, O) → U4(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
getleave_in(leaf(A), C, A, C) → getleave_out(leaf(A), C, A, C)
U4(A, B, C, L, O, getleave_out(A, tree(B, C), L, O)) → getleave_out(tree(A, B), C, L, O)
U1(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U2(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → U3(T1, T2, S1, S2, sameleaves_in(T, S))
sameleaves_in(leaf(L), leaf(L)) → sameleaves_out(leaf(L), leaf(L))
U3(T1, T2, S1, S2, sameleaves_out(T, S)) → sameleaves_out(tree(T1, T2), tree(S1, S2))

The argument filtering Pi contains the following mapping:
sameleaves_in(x1, x2) = sameleaves_in(x1, x2)
tree(x1, x2) = tree(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x4, x5)
getleave_in(x1, x2, x3, x4) = getleave_in(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
leaf(x1) = leaf(x1)
getleave_out(x1, x2, x3, x4) = getleave_out(x3, x4)
U2(x1, x2, x3, x4, x5, x6) = U2(x5, x6)
U3(x1, x2, x3, x4, x5) = U3(x5)
sameleaves_out(x1, x2) = sameleaves_out
U3¹(x1, x2, x3, x4, x5) = U3¹(x5)
U2¹(x1, x2, x3, x4, x5, x6) = U2¹(x5, x6)
U4¹(x1, x2, x3, x4, x5, x6) = U4¹(x6)
SAMELEAVES_IN(x1, x2) = SAMELEAVES_IN(x1, x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x4, x5)
GETLEAVE_IN(x1, x2, x3, x4) = GETLEAVE_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

GETLEAVE_IN(tree(A, B), C, L, O) → GETLEAVE_IN(A, tree(B, C), L, O)

The TRS R consists of the following rules:

sameleaves_in(tree(T1, T2), tree(S1, S2)) → U1(T1, T2, S1, S2, getleave_in(T1, T2, L, T))
getleave_in(tree(A, B), C, L, O) → U4(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
getleave_in(leaf(A), C, A, C) → getleave_out(leaf(A), C, A, C)
U4(A, B, C, L, O, getleave_out(A, tree(B, C), L, O)) → getleave_out(tree(A, B), C, L, O)
U1(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U2(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → U3(T1, T2, S1, S2, sameleaves_in(T, S))
sameleaves_in(leaf(L), leaf(L)) → sameleaves_out(leaf(L), leaf(L))
U3(T1, T2, S1, S2, sameleaves_out(T, S)) → sameleaves_out(tree(T1, T2), tree(S1, S2))

The argument filtering Pi contains the following mapping:
sameleaves_in(x1, x2) = sameleaves_in(x1, x2)
tree(x1, x2) = tree(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x4, x5)
getleave_in(x1, x2, x3, x4) = getleave_in(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
leaf(x1) = leaf(x1)
getleave_out(x1, x2, x3, x4) = getleave_out(x3, x4)
U2(x1, x2, x3, x4, x5, x6) = U2(x5, x6)
U3(x1, x2, x3, x4, x5) = U3(x5)
sameleaves_out(x1, x2) = sameleaves_out
GETLEAVE_IN(x1, x2, x3, x4) = GETLEAVE_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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

GETLEAVE_IN(tree(A, B), C, L, O) → GETLEAVE_IN(A, tree(B, C), L, O)

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2) = tree(x1, x2)
GETLEAVE_IN(x1, x2, x3, x4) = GETLEAVE_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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

GETLEAVE_IN(tree(A, B), C) → GETLEAVE_IN(A, tree(B, C))

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:

GETLEAVE_IN(tree(A, B), C) → GETLEAVE_IN(A, tree(B, C))
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

U1¹(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2¹(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U2¹(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → SAMELEAVES_IN(T, S)
SAMELEAVES_IN(tree(T1, T2), tree(S1, S2)) → U1¹(T1, T2, S1, S2, getleave_in(T1, T2, L, T))

The TRS R consists of the following rules:

sameleaves_in(tree(T1, T2), tree(S1, S2)) → U1(T1, T2, S1, S2, getleave_in(T1, T2, L, T))
getleave_in(tree(A, B), C, L, O) → U4(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
getleave_in(leaf(A), C, A, C) → getleave_out(leaf(A), C, A, C)
U4(A, B, C, L, O, getleave_out(A, tree(B, C), L, O)) → getleave_out(tree(A, B), C, L, O)
U1(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U2(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → U3(T1, T2, S1, S2, sameleaves_in(T, S))
sameleaves_in(leaf(L), leaf(L)) → sameleaves_out(leaf(L), leaf(L))
U3(T1, T2, S1, S2, sameleaves_out(T, S)) → sameleaves_out(tree(T1, T2), tree(S1, S2))

The argument filtering Pi contains the following mapping:
sameleaves_in(x1, x2) = sameleaves_in(x1, x2)
tree(x1, x2) = tree(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x3, x4, x5)
getleave_in(x1, x2, x3, x4) = getleave_in(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
leaf(x1) = leaf(x1)
getleave_out(x1, x2, x3, x4) = getleave_out(x3, x4)
U2(x1, x2, x3, x4, x5, x6) = U2(x5, x6)
U3(x1, x2, x3, x4, x5) = U3(x5)
sameleaves_out(x1, x2) = sameleaves_out
U2¹(x1, x2, x3, x4, x5, x6) = U2¹(x5, x6)
SAMELEAVES_IN(x1, x2) = SAMELEAVES_IN(x1, x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x4, x5)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

U1¹(T1, T2, S1, S2, getleave_out(T1, T2, L, T)) → U2¹(T1, T2, S1, S2, T, getleave_in(S1, S2, L, S))
U2¹(T1, T2, S1, S2, T, getleave_out(S1, S2, L, S)) → SAMELEAVES_IN(T, S)
SAMELEAVES_IN(tree(T1, T2), tree(S1, S2)) → U1¹(T1, T2, S1, S2, getleave_in(T1, T2, L, T))

The TRS R consists of the following rules:

getleave_in(tree(A, B), C, L, O) → U4(A, B, C, L, O, getleave_in(A, tree(B, C), L, O))
getleave_in(leaf(A), C, A, C) → getleave_out(leaf(A), C, A, C)
U4(A, B, C, L, O, getleave_out(A, tree(B, C), L, O)) → getleave_out(tree(A, B), C, L, O)

The argument filtering Pi contains the following mapping:
tree(x1, x2) = tree(x1, x2)
getleave_in(x1, x2, x3, x4) = getleave_in(x1, x2)
U4(x1, x2, x3, x4, x5, x6) = U4(x6)
leaf(x1) = leaf(x1)
getleave_out(x1, x2, x3, x4) = getleave_out(x3, x4)
U2¹(x1, x2, x3, x4, x5, x6) = U2¹(x5, x6)
SAMELEAVES_IN(x1, x2) = SAMELEAVES_IN(x1, x2)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x4, x5)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

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

SAMELEAVES_IN(tree(T1, T2), tree(S1, S2)) → U1¹(S1, S2, getleave_in(T1, T2))
U2¹(T, getleave_out(L, S)) → SAMELEAVES_IN(T, S)
U1¹(S1, S2, getleave_out(L, T)) → U2¹(T, getleave_in(S1, S2))

The TRS R consists of the following rules:

getleave_in(tree(A, B), C) → U4(getleave_in(A, tree(B, C)))
getleave_in(leaf(A), C) → getleave_out(A, C)
U4(getleave_out(L, O)) → getleave_out(L, O)

The set Q consists of the following terms:

getleave_in(x0, x1)
U4(x0)

We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

SAMELEAVES_IN(tree(T1, T2), tree(S1, S2)) → U1¹(S1, S2, getleave_in(T1, T2))
U2¹(T, getleave_out(L, S)) → SAMELEAVES_IN(T, S)
U1¹(S1, S2, getleave_out(L, T)) → U2¹(T, getleave_in(S1, S2))

The following rules are removed from R:

getleave_in(leaf(A), C) → getleave_out(A, C)

Used ordering: POLO with Polynomial interpretation [25]:

POL(SAMELEAVES_IN(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U1¹(x₁, x₂, x₃)) = 1 + 2·x₁ + 2·x₂ + 2·x₃
POL(U2¹(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U4(x₁)) = x₁
POL(getleave_in(x₁, x₂)) = 1 + x₁ + x₂
POL(getleave_out(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(leaf(x₁)) = 2·x₁
POL(tree(x₁, x₂)) = 2 + x₁ + x₂

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

getleave_in(tree(A, B), C) → U4(getleave_in(A, tree(B, C)))
U4(getleave_out(L, O)) → getleave_out(L, O)

The set Q consists of the following terms:

getleave_in(x0, x1)
U4(x0)

We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.