Left Termination of the query pattern reverse_concatenate_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

concatenate([], L, L).
concatenate(.(X, L1), L2, .(X, L3)) :- concatenate(L1, L2, L3).
member(X, .(X, L)).
member(X, .(Y, L)) :- member(X, L).
reverse(L, L1) :- reverse_concatenate(L, [], L1).
reverse_concatenate([], L, L).
reverse_concatenate(.(X, L1), L2, L3) :- reverse_concatenate(L1, .(X, L2), L3).

Queries:

reverse_concatenate(g,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:

reverse_concatenate_in(.(X, L1), L2, L3) → U4(X, L1, L2, L3, reverse_concatenate_in(L1, .(X, L2), L3))
reverse_concatenate_in([], L, L) → reverse_concatenate_out([], L, L)
U4(X, L1, L2, L3, reverse_concatenate_out(L1, .(X, L2), L3)) → reverse_concatenate_out(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in(x1, x2, x3)  =  reverse_concatenate_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
[]  =  []
reverse_concatenate_out(x1, x2, x3)  =  reverse_concatenate_out(x3)

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:

reverse_concatenate_in(.(X, L1), L2, L3) → U4(X, L1, L2, L3, reverse_concatenate_in(L1, .(X, L2), L3))
reverse_concatenate_in([], L, L) → reverse_concatenate_out([], L, L)
U4(X, L1, L2, L3, reverse_concatenate_out(L1, .(X, L2), L3)) → reverse_concatenate_out(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in(x1, x2, x3)  =  reverse_concatenate_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
[]  =  []
reverse_concatenate_out(x1, x2, x3)  =  reverse_concatenate_out(x3)


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

REVERSE_CONCATENATE_IN(.(X, L1), L2, L3) → U41(X, L1, L2, L3, reverse_concatenate_in(L1, .(X, L2), L3))
REVERSE_CONCATENATE_IN(.(X, L1), L2, L3) → REVERSE_CONCATENATE_IN(L1, .(X, L2), L3)

The TRS R consists of the following rules:

reverse_concatenate_in(.(X, L1), L2, L3) → U4(X, L1, L2, L3, reverse_concatenate_in(L1, .(X, L2), L3))
reverse_concatenate_in([], L, L) → reverse_concatenate_out([], L, L)
U4(X, L1, L2, L3, reverse_concatenate_out(L1, .(X, L2), L3)) → reverse_concatenate_out(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in(x1, x2, x3)  =  reverse_concatenate_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
[]  =  []
reverse_concatenate_out(x1, x2, x3)  =  reverse_concatenate_out(x3)
U41(x1, x2, x3, x4, x5)  =  U41(x5)
REVERSE_CONCATENATE_IN(x1, x2, x3)  =  REVERSE_CONCATENATE_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:

REVERSE_CONCATENATE_IN(.(X, L1), L2, L3) → U41(X, L1, L2, L3, reverse_concatenate_in(L1, .(X, L2), L3))
REVERSE_CONCATENATE_IN(.(X, L1), L2, L3) → REVERSE_CONCATENATE_IN(L1, .(X, L2), L3)

The TRS R consists of the following rules:

reverse_concatenate_in(.(X, L1), L2, L3) → U4(X, L1, L2, L3, reverse_concatenate_in(L1, .(X, L2), L3))
reverse_concatenate_in([], L, L) → reverse_concatenate_out([], L, L)
U4(X, L1, L2, L3, reverse_concatenate_out(L1, .(X, L2), L3)) → reverse_concatenate_out(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in(x1, x2, x3)  =  reverse_concatenate_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
[]  =  []
reverse_concatenate_out(x1, x2, x3)  =  reverse_concatenate_out(x3)
U41(x1, x2, x3, x4, x5)  =  U41(x5)
REVERSE_CONCATENATE_IN(x1, x2, x3)  =  REVERSE_CONCATENATE_IN(x1, x2)

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:

REVERSE_CONCATENATE_IN(.(X, L1), L2, L3) → REVERSE_CONCATENATE_IN(L1, .(X, L2), L3)

The TRS R consists of the following rules:

reverse_concatenate_in(.(X, L1), L2, L3) → U4(X, L1, L2, L3, reverse_concatenate_in(L1, .(X, L2), L3))
reverse_concatenate_in([], L, L) → reverse_concatenate_out([], L, L)
U4(X, L1, L2, L3, reverse_concatenate_out(L1, .(X, L2), L3)) → reverse_concatenate_out(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in(x1, x2, x3)  =  reverse_concatenate_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
[]  =  []
reverse_concatenate_out(x1, x2, x3)  =  reverse_concatenate_out(x3)
REVERSE_CONCATENATE_IN(x1, x2, x3)  =  REVERSE_CONCATENATE_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

REVERSE_CONCATENATE_IN(.(X, L1), L2, L3) → REVERSE_CONCATENATE_IN(L1, .(X, L2), L3)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
REVERSE_CONCATENATE_IN(x1, x2, x3)  =  REVERSE_CONCATENATE_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

REVERSE_CONCATENATE_IN(.(X, L1), L2) → REVERSE_CONCATENATE_IN(L1, .(X, L2))

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: