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:

app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
app([], Ys, Ys).
reverse(.(X, Xs), Ys) :- ','(reverse(Xs, Zs), app(Zs, .(X, []), Ys)).
reverse([], []).

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([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
[]  =  []
reverse_out(x1, x2)  =  reverse_out(x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
U3(x1, x2, x3, x4)  =  U3(x4)
app_in(x1, x2, x3)  =  app_in(x1, x2)
app_out(x1, x2, x3)  =  app_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(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:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
[]  =  []
reverse_out(x1, x2)  =  reverse_out(x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
U3(x1, x2, x3, x4)  =  U3(x4)
app_in(x1, x2, x3)  =  app_in(x1, x2)
app_out(x1, x2, x3)  =  app_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(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:

REVERSE_IN(.(X, Xs), Ys) → U21(X, Xs, Ys, reverse_in(Xs, Zs))
REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)
U21(X, Xs, Ys, reverse_out(Xs, Zs)) → U31(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
U21(X, Xs, Ys, reverse_out(Xs, Zs)) → APP_IN(Zs, .(X, []), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
[]  =  []
reverse_out(x1, x2)  =  reverse_out(x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
U3(x1, x2, x3, x4)  =  U3(x4)
app_in(x1, x2, x3)  =  app_in(x1, x2)
app_out(x1, x2, x3)  =  app_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
REVERSE_IN(x1, x2)  =  REVERSE_IN(x1)
U31(x1, x2, x3, x4)  =  U31(x4)
U21(x1, x2, x3, x4)  =  U21(x1, x4)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)
APP_IN(x1, x2, x3)  =  APP_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_IN(.(X, Xs), Ys) → U21(X, Xs, Ys, reverse_in(Xs, Zs))
REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)
U21(X, Xs, Ys, reverse_out(Xs, Zs)) → U31(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
U21(X, Xs, Ys, reverse_out(Xs, Zs)) → APP_IN(Zs, .(X, []), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
[]  =  []
reverse_out(x1, x2)  =  reverse_out(x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
U3(x1, x2, x3, x4)  =  U3(x4)
app_in(x1, x2, x3)  =  app_in(x1, x2)
app_out(x1, x2, x3)  =  app_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
REVERSE_IN(x1, x2)  =  REVERSE_IN(x1)
U31(x1, x2, x3, x4)  =  U31(x4)
U21(x1, x2, x3, x4)  =  U21(x1, x4)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)
APP_IN(x1, x2, x3)  =  APP_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

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

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
[]  =  []
reverse_out(x1, x2)  =  reverse_out(x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
U3(x1, x2, x3, x4)  =  U3(x4)
app_in(x1, x2, x3)  =  app_in(x1, x2)
app_out(x1, x2, x3)  =  app_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
APP_IN(x1, x2, x3)  =  APP_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

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

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

APP_IN(.(X, Xs), Ys) → APP_IN(Xs, Ys)

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: 



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

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

REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)

The TRS R consists of the following rules:

reverse_in([], []) → reverse_out([], [])
reverse_in(.(X, Xs), Ys) → U2(X, Xs, Ys, reverse_in(Xs, Zs))
U2(X, Xs, Ys, reverse_out(Xs, Zs)) → U3(X, Xs, Ys, app_in(Zs, .(X, []), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Xs, Ys, app_out(Zs, .(X, []), Ys)) → reverse_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
reverse_in(x1, x2)  =  reverse_in(x1)
[]  =  []
reverse_out(x1, x2)  =  reverse_out(x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4)  =  U2(x1, x4)
U3(x1, x2, x3, x4)  =  U3(x4)
app_in(x1, x2, x3)  =  app_in(x1, x2)
app_out(x1, x2, x3)  =  app_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
REVERSE_IN(x1, x2)  =  REVERSE_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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

REVERSE_IN(.(X, Xs), Ys) → REVERSE_IN(Xs, Zs)

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

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

REVERSE_IN(.(X, Xs)) → REVERSE_IN(Xs)

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: