Left Termination of the query pattern reverse_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

reverse(X1s, X2s) :- reverse(X1s, [], X2s).
reverse([], Xs, Xs).
reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).

Queries:

reverse(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_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
U1(x1, x2, x3)  =  U1(x3)
reverse_in(x1, x2, x3)  =  reverse_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
reverse_out(x1, x2, x3)  =  reverse_out(x3)
reverse_out(x1, x2)  =  reverse_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:

reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
U1(x1, x2, x3)  =  U1(x3)
reverse_in(x1, x2, x3)  =  reverse_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
reverse_out(x1, x2, x3)  =  reverse_out(x3)
reverse_out(x1, x2)  =  reverse_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:

REVERSE_IN(X1s, X2s) → U11(X1s, X2s, reverse_in(X1s, [], X2s))
REVERSE_IN(X1s, X2s) → REVERSE_IN(X1s, [], X2s)
REVERSE_IN(.(X, X1s), X2s, Ys) → U21(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
U1(x1, x2, x3)  =  U1(x3)
reverse_in(x1, x2, x3)  =  reverse_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
reverse_out(x1, x2, x3)  =  reverse_out(x3)
reverse_out(x1, x2)  =  reverse_out(x2)
REVERSE_IN(x1, x2)  =  REVERSE_IN(x1)
REVERSE_IN(x1, x2, x3)  =  REVERSE_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3)  =  U11(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:

REVERSE_IN(X1s, X2s) → U11(X1s, X2s, reverse_in(X1s, [], X2s))
REVERSE_IN(X1s, X2s) → REVERSE_IN(X1s, [], X2s)
REVERSE_IN(.(X, X1s), X2s, Ys) → U21(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
U1(x1, x2, x3)  =  U1(x3)
reverse_in(x1, x2, x3)  =  reverse_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
reverse_out(x1, x2, x3)  =  reverse_out(x3)
reverse_out(x1, x2)  =  reverse_out(x2)
REVERSE_IN(x1, x2)  =  REVERSE_IN(x1)
REVERSE_IN(x1, x2, x3)  =  REVERSE_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3)  =  U11(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

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

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

REVERSE_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in(X1s, X2s) → U1(X1s, X2s, reverse_in(X1s, [], X2s))
reverse_in(.(X, X1s), X2s, Ys) → U2(X, X1s, X2s, Ys, reverse_in(X1s, .(X, X2s), Ys))
reverse_in([], Xs, Xs) → reverse_out([], Xs, Xs)
U2(X, X1s, X2s, Ys, reverse_out(X1s, .(X, X2s), Ys)) → reverse_out(.(X, X1s), X2s, Ys)
U1(X1s, X2s, reverse_out(X1s, [], X2s)) → reverse_out(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
U1(x1, x2, x3)  =  U1(x3)
reverse_in(x1, x2, x3)  =  reverse_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
reverse_out(x1, x2, x3)  =  reverse_out(x3)
reverse_out(x1, x2)  =  reverse_out(x2)
REVERSE_IN(x1, x2, x3)  =  REVERSE_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_IN(.(X, X1s), X2s, Ys) → REVERSE_IN(X1s, .(X, X2s), Ys)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
REVERSE_IN(x1, x2, x3)  =  REVERSE_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_IN(.(X, X1s), X2s) → REVERSE_IN(X1s, .(X, X2s))

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: