Left Termination of the query pattern flat_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

flat(niltree, nil).
flat(tree(X, niltree, T), cons(X, Xs)) :- flat(T, Xs).
flat(tree(X, tree(Y, T1, T2), T3), Xs) :- flat(tree(Y, T1, tree(X, T2, T3)), Xs).

Queries:

flat(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:

flat_in(tree(X, tree(Y, T1, T2), T3), Xs) → U2(X, Y, T1, T2, T3, Xs, flat_in(tree(Y, T1, tree(X, T2, T3)), Xs))
flat_in(tree(X, niltree, T), cons(X, Xs)) → U1(X, T, Xs, flat_in(T, Xs))
flat_in(niltree, nil) → flat_out(niltree, nil)
U1(X, T, Xs, flat_out(T, Xs)) → flat_out(tree(X, niltree, T), cons(X, Xs))
U2(X, Y, T1, T2, T3, Xs, flat_out(tree(Y, T1, tree(X, T2, T3)), Xs)) → flat_out(tree(X, tree(Y, T1, T2), T3), Xs)

The argument filtering Pi contains the following mapping:
flat_in(x1, x2)  =  flat_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7)  =  U2(x7)
niltree  =  niltree
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
nil  =  nil
flat_out(x1, x2)  =  flat_out(x2)

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:

flat_in(tree(X, tree(Y, T1, T2), T3), Xs) → U2(X, Y, T1, T2, T3, Xs, flat_in(tree(Y, T1, tree(X, T2, T3)), Xs))
flat_in(tree(X, niltree, T), cons(X, Xs)) → U1(X, T, Xs, flat_in(T, Xs))
flat_in(niltree, nil) → flat_out(niltree, nil)
U1(X, T, Xs, flat_out(T, Xs)) → flat_out(tree(X, niltree, T), cons(X, Xs))
U2(X, Y, T1, T2, T3, Xs, flat_out(tree(Y, T1, tree(X, T2, T3)), Xs)) → flat_out(tree(X, tree(Y, T1, T2), T3), Xs)

The argument filtering Pi contains the following mapping:
flat_in(x1, x2)  =  flat_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7)  =  U2(x7)
niltree  =  niltree
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
nil  =  nil
flat_out(x1, x2)  =  flat_out(x2)


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

FLAT_IN(tree(X, tree(Y, T1, T2), T3), Xs) → U21(X, Y, T1, T2, T3, Xs, flat_in(tree(Y, T1, tree(X, T2, T3)), Xs))
FLAT_IN(tree(X, tree(Y, T1, T2), T3), Xs) → FLAT_IN(tree(Y, T1, tree(X, T2, T3)), Xs)
FLAT_IN(tree(X, niltree, T), cons(X, Xs)) → U11(X, T, Xs, flat_in(T, Xs))
FLAT_IN(tree(X, niltree, T), cons(X, Xs)) → FLAT_IN(T, Xs)

The TRS R consists of the following rules:

flat_in(tree(X, tree(Y, T1, T2), T3), Xs) → U2(X, Y, T1, T2, T3, Xs, flat_in(tree(Y, T1, tree(X, T2, T3)), Xs))
flat_in(tree(X, niltree, T), cons(X, Xs)) → U1(X, T, Xs, flat_in(T, Xs))
flat_in(niltree, nil) → flat_out(niltree, nil)
U1(X, T, Xs, flat_out(T, Xs)) → flat_out(tree(X, niltree, T), cons(X, Xs))
U2(X, Y, T1, T2, T3, Xs, flat_out(tree(Y, T1, tree(X, T2, T3)), Xs)) → flat_out(tree(X, tree(Y, T1, T2), T3), Xs)

The argument filtering Pi contains the following mapping:
flat_in(x1, x2)  =  flat_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7)  =  U2(x7)
niltree  =  niltree
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
nil  =  nil
flat_out(x1, x2)  =  flat_out(x2)
U21(x1, x2, x3, x4, x5, x6, x7)  =  U21(x7)
U11(x1, x2, x3, x4)  =  U11(x1, x4)
FLAT_IN(x1, x2)  =  FLAT_IN(x1)

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:

FLAT_IN(tree(X, tree(Y, T1, T2), T3), Xs) → U21(X, Y, T1, T2, T3, Xs, flat_in(tree(Y, T1, tree(X, T2, T3)), Xs))
FLAT_IN(tree(X, tree(Y, T1, T2), T3), Xs) → FLAT_IN(tree(Y, T1, tree(X, T2, T3)), Xs)
FLAT_IN(tree(X, niltree, T), cons(X, Xs)) → U11(X, T, Xs, flat_in(T, Xs))
FLAT_IN(tree(X, niltree, T), cons(X, Xs)) → FLAT_IN(T, Xs)

The TRS R consists of the following rules:

flat_in(tree(X, tree(Y, T1, T2), T3), Xs) → U2(X, Y, T1, T2, T3, Xs, flat_in(tree(Y, T1, tree(X, T2, T3)), Xs))
flat_in(tree(X, niltree, T), cons(X, Xs)) → U1(X, T, Xs, flat_in(T, Xs))
flat_in(niltree, nil) → flat_out(niltree, nil)
U1(X, T, Xs, flat_out(T, Xs)) → flat_out(tree(X, niltree, T), cons(X, Xs))
U2(X, Y, T1, T2, T3, Xs, flat_out(tree(Y, T1, tree(X, T2, T3)), Xs)) → flat_out(tree(X, tree(Y, T1, T2), T3), Xs)

The argument filtering Pi contains the following mapping:
flat_in(x1, x2)  =  flat_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7)  =  U2(x7)
niltree  =  niltree
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
nil  =  nil
flat_out(x1, x2)  =  flat_out(x2)
U21(x1, x2, x3, x4, x5, x6, x7)  =  U21(x7)
U11(x1, x2, x3, x4)  =  U11(x1, x4)
FLAT_IN(x1, x2)  =  FLAT_IN(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

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

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

FLAT_IN(tree(X, tree(Y, T1, T2), T3), Xs) → FLAT_IN(tree(Y, T1, tree(X, T2, T3)), Xs)
FLAT_IN(tree(X, niltree, T), cons(X, Xs)) → FLAT_IN(T, Xs)

The TRS R consists of the following rules:

flat_in(tree(X, tree(Y, T1, T2), T3), Xs) → U2(X, Y, T1, T2, T3, Xs, flat_in(tree(Y, T1, tree(X, T2, T3)), Xs))
flat_in(tree(X, niltree, T), cons(X, Xs)) → U1(X, T, Xs, flat_in(T, Xs))
flat_in(niltree, nil) → flat_out(niltree, nil)
U1(X, T, Xs, flat_out(T, Xs)) → flat_out(tree(X, niltree, T), cons(X, Xs))
U2(X, Y, T1, T2, T3, Xs, flat_out(tree(Y, T1, tree(X, T2, T3)), Xs)) → flat_out(tree(X, tree(Y, T1, T2), T3), Xs)

The argument filtering Pi contains the following mapping:
flat_in(x1, x2)  =  flat_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5, x6, x7)  =  U2(x7)
niltree  =  niltree
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
nil  =  nil
flat_out(x1, x2)  =  flat_out(x2)
FLAT_IN(x1, x2)  =  FLAT_IN(x1)

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:

FLAT_IN(tree(X, tree(Y, T1, T2), T3), Xs) → FLAT_IN(tree(Y, T1, tree(X, T2, T3)), Xs)
FLAT_IN(tree(X, niltree, T), cons(X, Xs)) → FLAT_IN(T, Xs)

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

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
                      ↳ UsableRulesReductionPairsProof

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

FLAT_IN(tree(X, tree(Y, T1, T2), T3)) → FLAT_IN(tree(Y, T1, tree(X, T2, T3)))
FLAT_IN(tree(X, niltree, T)) → FLAT_IN(T)

R is empty.
Q is empty.
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:

FLAT_IN(tree(X, niltree, T)) → FLAT_IN(T)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(FLAT_IN(x1)) = 2·x1   
POL(niltree) = 0   
POL(tree(x1, x2, x3)) = x1 + 2·x2 + x3   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof
                        ↳ QDP
                          ↳ RuleRemovalProof

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

FLAT_IN(tree(X, tree(Y, T1, T2), T3)) → FLAT_IN(tree(Y, T1, tree(X, T2, T3)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

FLAT_IN(tree(X, tree(Y, T1, T2), T3)) → FLAT_IN(tree(Y, T1, tree(X, T2, T3)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(FLAT_IN(x1)) = 2·x1   
POL(tree(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + x3   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof
                        ↳ QDP
                          ↳ RuleRemovalProof
                            ↳ QDP
                              ↳ PisEmptyProof

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