Left Termination of the query pattern flatten_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

flatten(atom(X), .(X, [])).
flatten(cons(atom(X), U), .(X, Y)) :- flatten(U, Y).
flatten(cons(cons(U, V), W), X) :- flatten(cons(U, cons(V, W)), X).

Queries:

flatten(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:

flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
flatten_in(x1, x2)  =  flatten_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
atom(x1)  =  atom(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
flatten_out(x1, x2)  =  flatten_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:

flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
flatten_in(x1, x2)  =  flatten_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
atom(x1)  =  atom(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
flatten_out(x1, x2)  =  flatten_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:

FLATTEN_IN(cons(cons(U, V), W), X) → U21(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
FLATTEN_IN(cons(cons(U, V), W), X) → FLATTEN_IN(cons(U, cons(V, W)), X)
FLATTEN_IN(cons(atom(X), U), .(X, Y)) → U11(X, U, Y, flatten_in(U, Y))
FLATTEN_IN(cons(atom(X), U), .(X, Y)) → FLATTEN_IN(U, Y)

The TRS R consists of the following rules:

flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
flatten_in(x1, x2)  =  flatten_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
atom(x1)  =  atom(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
flatten_out(x1, x2)  =  flatten_out(x2)
FLATTEN_IN(x1, x2)  =  FLATTEN_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x1, x4)

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:

FLATTEN_IN(cons(cons(U, V), W), X) → U21(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
FLATTEN_IN(cons(cons(U, V), W), X) → FLATTEN_IN(cons(U, cons(V, W)), X)
FLATTEN_IN(cons(atom(X), U), .(X, Y)) → U11(X, U, Y, flatten_in(U, Y))
FLATTEN_IN(cons(atom(X), U), .(X, Y)) → FLATTEN_IN(U, Y)

The TRS R consists of the following rules:

flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
flatten_in(x1, x2)  =  flatten_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
atom(x1)  =  atom(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
flatten_out(x1, x2)  =  flatten_out(x2)
FLATTEN_IN(x1, x2)  =  FLATTEN_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x1, x4)

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:

FLATTEN_IN(cons(atom(X), U), .(X, Y)) → FLATTEN_IN(U, Y)
FLATTEN_IN(cons(cons(U, V), W), X) → FLATTEN_IN(cons(U, cons(V, W)), X)

The TRS R consists of the following rules:

flatten_in(cons(cons(U, V), W), X) → U2(U, V, W, X, flatten_in(cons(U, cons(V, W)), X))
flatten_in(cons(atom(X), U), .(X, Y)) → U1(X, U, Y, flatten_in(U, Y))
flatten_in(atom(X), .(X, [])) → flatten_out(atom(X), .(X, []))
U1(X, U, Y, flatten_out(U, Y)) → flatten_out(cons(atom(X), U), .(X, Y))
U2(U, V, W, X, flatten_out(cons(U, cons(V, W)), X)) → flatten_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
flatten_in(x1, x2)  =  flatten_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
atom(x1)  =  atom(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
[]  =  []
flatten_out(x1, x2)  =  flatten_out(x2)
FLATTEN_IN(x1, x2)  =  FLATTEN_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:

FLATTEN_IN(cons(atom(X), U), .(X, Y)) → FLATTEN_IN(U, Y)
FLATTEN_IN(cons(cons(U, V), W), X) → FLATTEN_IN(cons(U, cons(V, W)), X)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2)  =  cons(x1, x2)
atom(x1)  =  atom(x1)
.(x1, x2)  =  .(x1, x2)
FLATTEN_IN(x1, x2)  =  FLATTEN_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:

FLATTEN_IN(cons(cons(U, V), W)) → FLATTEN_IN(cons(U, cons(V, W)))
FLATTEN_IN(cons(atom(X), U)) → FLATTEN_IN(U)

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:

FLATTEN_IN(cons(cons(U, V), W)) → FLATTEN_IN(cons(U, cons(V, W)))
FLATTEN_IN(cons(atom(X), U)) → FLATTEN_IN(U)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(FLATTEN_IN(x1)) = 2·x1   
POL(atom(x1)) = x1   
POL(cons(x1, x2)) = 2 + 2·x1 + x2   



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