Left Termination of the query pattern reverse_in_3(g, a, g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

reverse([], X, X).
reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).

Queries:

reverse(g,a,g).

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(.(X, Y), Z, U) → U1(X, Y, Z, U, reverse_in(Y, Z, .(X, U)))
reverse_in([], X, X) → reverse_out([], X, X)
U1(X, Y, Z, U, reverse_out(Y, Z, .(X, U))) → reverse_out(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2, x3)  =  reverse_in(x1, x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
reverse_out(x1, x2, x3)  =  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(.(X, Y), Z, U) → U1(X, Y, Z, U, reverse_in(Y, Z, .(X, U)))
reverse_in([], X, X) → reverse_out([], X, X)
U1(X, Y, Z, U, reverse_out(Y, Z, .(X, U))) → reverse_out(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2, x3)  =  reverse_in(x1, x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
reverse_out(x1, x2, x3)  =  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(.(X, Y), Z, U) → U11(X, Y, Z, U, reverse_in(Y, Z, .(X, U)))
REVERSE_IN(.(X, Y), Z, U) → REVERSE_IN(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in(.(X, Y), Z, U) → U1(X, Y, Z, U, reverse_in(Y, Z, .(X, U)))
reverse_in([], X, X) → reverse_out([], X, X)
U1(X, Y, Z, U, reverse_out(Y, Z, .(X, U))) → reverse_out(.(X, Y), Z, U)

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

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(.(X, Y), Z, U) → U11(X, Y, Z, U, reverse_in(Y, Z, .(X, U)))
REVERSE_IN(.(X, Y), Z, U) → REVERSE_IN(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in(.(X, Y), Z, U) → U1(X, Y, Z, U, reverse_in(Y, Z, .(X, U)))
reverse_in([], X, X) → reverse_out([], X, X)
U1(X, Y, Z, U, reverse_out(Y, Z, .(X, U))) → reverse_out(.(X, Y), Z, U)

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

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_IN(.(X, Y), Z, U) → REVERSE_IN(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in(.(X, Y), Z, U) → U1(X, Y, Z, U, reverse_in(Y, Z, .(X, U)))
reverse_in([], X, X) → reverse_out([], X, X)
U1(X, Y, Z, U, reverse_out(Y, Z, .(X, U))) → reverse_out(.(X, Y), Z, U)

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

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, Y), Z, U) → REVERSE_IN(Y, Z, .(X, U))

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

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, Y), U) → REVERSE_IN(Y, .(X, U))

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: