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([], []).
flat(.([], T), R) :- flat(T, R).
flat(.(.(H, T), TT), .(H, R)) :- flat(.(T, TT), R).

Queries:

flat(g,a).

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

flat_in_ga([], []) → flat_out_ga([], [])
flat_in_ga(.([], T), R) → U1_ga(T, R, flat_in_ga(T, R))
flat_in_ga(.(.(H, T), TT), .(H, R)) → U2_ga(H, T, TT, R, flat_in_ga(.(T, TT), R))
U2_ga(H, T, TT, R, flat_out_ga(.(T, TT), R)) → flat_out_ga(.(.(H, T), TT), .(H, R))
U1_ga(T, R, flat_out_ga(T, R)) → flat_out_ga(.([], T), R)

The argument filtering Pi contains the following mapping:
flat_in_ga(x1, x2)  =  flat_in_ga(x1)
[]  =  []
flat_out_ga(x1, x2)  =  flat_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(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:

flat_in_ga([], []) → flat_out_ga([], [])
flat_in_ga(.([], T), R) → U1_ga(T, R, flat_in_ga(T, R))
flat_in_ga(.(.(H, T), TT), .(H, R)) → U2_ga(H, T, TT, R, flat_in_ga(.(T, TT), R))
U2_ga(H, T, TT, R, flat_out_ga(.(T, TT), R)) → flat_out_ga(.(.(H, T), TT), .(H, R))
U1_ga(T, R, flat_out_ga(T, R)) → flat_out_ga(.([], T), R)

The argument filtering Pi contains the following mapping:
flat_in_ga(x1, x2)  =  flat_in_ga(x1)
[]  =  []
flat_out_ga(x1, x2)  =  flat_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(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:

FLAT_IN_GA(.([], T), R) → U1_GA(T, R, flat_in_ga(T, R))
FLAT_IN_GA(.([], T), R) → FLAT_IN_GA(T, R)
FLAT_IN_GA(.(.(H, T), TT), .(H, R)) → U2_GA(H, T, TT, R, flat_in_ga(.(T, TT), R))
FLAT_IN_GA(.(.(H, T), TT), .(H, R)) → FLAT_IN_GA(.(T, TT), R)

The TRS R consists of the following rules:

flat_in_ga([], []) → flat_out_ga([], [])
flat_in_ga(.([], T), R) → U1_ga(T, R, flat_in_ga(T, R))
flat_in_ga(.(.(H, T), TT), .(H, R)) → U2_ga(H, T, TT, R, flat_in_ga(.(T, TT), R))
U2_ga(H, T, TT, R, flat_out_ga(.(T, TT), R)) → flat_out_ga(.(.(H, T), TT), .(H, R))
U1_ga(T, R, flat_out_ga(T, R)) → flat_out_ga(.([], T), R)

The argument filtering Pi contains the following mapping:
flat_in_ga(x1, x2)  =  flat_in_ga(x1)
[]  =  []
flat_out_ga(x1, x2)  =  flat_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x5)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x5)
FLAT_IN_GA(x1, x2)  =  FLAT_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)

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_GA(.([], T), R) → U1_GA(T, R, flat_in_ga(T, R))
FLAT_IN_GA(.([], T), R) → FLAT_IN_GA(T, R)
FLAT_IN_GA(.(.(H, T), TT), .(H, R)) → U2_GA(H, T, TT, R, flat_in_ga(.(T, TT), R))
FLAT_IN_GA(.(.(H, T), TT), .(H, R)) → FLAT_IN_GA(.(T, TT), R)

The TRS R consists of the following rules:

flat_in_ga([], []) → flat_out_ga([], [])
flat_in_ga(.([], T), R) → U1_ga(T, R, flat_in_ga(T, R))
flat_in_ga(.(.(H, T), TT), .(H, R)) → U2_ga(H, T, TT, R, flat_in_ga(.(T, TT), R))
U2_ga(H, T, TT, R, flat_out_ga(.(T, TT), R)) → flat_out_ga(.(.(H, T), TT), .(H, R))
U1_ga(T, R, flat_out_ga(T, R)) → flat_out_ga(.([], T), R)

The argument filtering Pi contains the following mapping:
flat_in_ga(x1, x2)  =  flat_in_ga(x1)
[]  =  []
flat_out_ga(x1, x2)  =  flat_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x5)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x5)
FLAT_IN_GA(x1, x2)  =  FLAT_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x3)

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_GA(.([], T), R) → FLAT_IN_GA(T, R)
FLAT_IN_GA(.(.(H, T), TT), .(H, R)) → FLAT_IN_GA(.(T, TT), R)

The TRS R consists of the following rules:

flat_in_ga([], []) → flat_out_ga([], [])
flat_in_ga(.([], T), R) → U1_ga(T, R, flat_in_ga(T, R))
flat_in_ga(.(.(H, T), TT), .(H, R)) → U2_ga(H, T, TT, R, flat_in_ga(.(T, TT), R))
U2_ga(H, T, TT, R, flat_out_ga(.(T, TT), R)) → flat_out_ga(.(.(H, T), TT), .(H, R))
U1_ga(T, R, flat_out_ga(T, R)) → flat_out_ga(.([], T), R)

The argument filtering Pi contains the following mapping:
flat_in_ga(x1, x2)  =  flat_in_ga(x1)
[]  =  []
flat_out_ga(x1, x2)  =  flat_out_ga(x2)
.(x1, x2)  =  .(x1, x2)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x5)
FLAT_IN_GA(x1, x2)  =  FLAT_IN_GA(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_GA(.([], T), R) → FLAT_IN_GA(T, R)
FLAT_IN_GA(.(.(H, T), TT), .(H, R)) → FLAT_IN_GA(.(T, TT), R)

R is empty.
The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x1, x2)
FLAT_IN_GA(x1, x2)  =  FLAT_IN_GA(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_GA(.(.(H, T), TT)) → FLAT_IN_GA(.(T, TT))
FLAT_IN_GA(.([], T)) → FLAT_IN_GA(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_GA(.([], T)) → FLAT_IN_GA(T)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 2·x1 + 2·x2   
POL(FLAT_IN_GA(x1)) = 2·x1   
POL([]) = 0   



↳ 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_GA(.(.(H, T), TT)) → FLAT_IN_GA(.(T, TT))

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_GA(.(.(H, T), TT)) → FLAT_IN_GA(.(T, TT))


Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(FLAT_IN_GA(x1)) = 2·x1   



↳ 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.