Left Termination of the query pattern suffix_in_2(g, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

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

Queries:

suffix(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:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

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

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

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

SUFFIX_IN(Xs, Ys) → U11(Xs, Ys, app_in(X, Xs, Ys))
SUFFIX_IN(Xs, Ys) → APP_IN(X, Xs, Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U21(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:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_in(x1, x2)  =  suffix_in(x1)
U1(x1, x2, x3)  =  U1(x3)
app_in(x1, x2, x3)  =  app_in(x2)
.(x1, x2)  =  .(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x1, x3)
suffix_out(x1, x2)  =  suffix_out(x2)
SUFFIX_IN(x1, x2)  =  SUFFIX_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3)  =  U11(x3)
APP_IN(x1, x2, x3)  =  APP_IN(x2)

We have to consider all (P,R,Pi)-chains

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

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

SUFFIX_IN(Xs, Ys) → U11(Xs, Ys, app_in(X, Xs, Ys))
SUFFIX_IN(Xs, Ys) → APP_IN(X, Xs, Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U21(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:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_in(x1, x2)  =  suffix_in(x1)
U1(x1, x2, x3)  =  U1(x3)
app_in(x1, x2, x3)  =  app_in(x2)
.(x1, x2)  =  .(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x1, x3)
suffix_out(x1, x2)  =  suffix_out(x2)
SUFFIX_IN(x1, x2)  =  SUFFIX_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3)  =  U11(x3)
APP_IN(x1, x2, x3)  =  APP_IN(x2)

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
  ↳ PrologToPiTRSProof

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:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_in(x1, x2)  =  suffix_in(x1)
U1(x1, x2, x3)  =  U1(x3)
app_in(x1, x2, x3)  =  app_in(x2)
.(x1, x2)  =  .(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x1, x3)
suffix_out(x1, x2)  =  suffix_out(x2)
APP_IN(x1, x2, x3)  =  APP_IN(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
  ↳ PrologToPiTRSProof

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)  =  .(x2)
APP_IN(x1, x2, x3)  =  APP_IN(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
                      ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

APP_IN(Ys) → APP_IN(Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_IN(Ys) → APP_IN(Ys)

The TRS R consists of the following rules:none


s = APP_IN(Ys) evaluates to t =APP_IN(Ys)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_IN(Ys) to APP_IN(Ys).




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:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

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

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

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

SUFFIX_IN(Xs, Ys) → U11(Xs, Ys, app_in(X, Xs, Ys))
SUFFIX_IN(Xs, Ys) → APP_IN(X, Xs, Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U21(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:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_in(x1, x2)  =  suffix_in(x1)
U1(x1, x2, x3)  =  U1(x1, x3)
app_in(x1, x2, x3)  =  app_in(x2)
.(x1, x2)  =  .(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x1, x2, x3)
suffix_out(x1, x2)  =  suffix_out(x1, x2)
SUFFIX_IN(x1, x2)  =  SUFFIX_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x3, x5)
U11(x1, x2, x3)  =  U11(x1, x3)
APP_IN(x1, x2, x3)  =  APP_IN(x2)

We have to consider all (P,R,Pi)-chains

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

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

SUFFIX_IN(Xs, Ys) → U11(Xs, Ys, app_in(X, Xs, Ys))
SUFFIX_IN(Xs, Ys) → APP_IN(X, Xs, Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U21(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:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_in(x1, x2)  =  suffix_in(x1)
U1(x1, x2, x3)  =  U1(x1, x3)
app_in(x1, x2, x3)  =  app_in(x2)
.(x1, x2)  =  .(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x1, x2, x3)
suffix_out(x1, x2)  =  suffix_out(x1, x2)
SUFFIX_IN(x1, x2)  =  SUFFIX_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x3, x5)
U11(x1, x2, x3)  =  U11(x1, x3)
APP_IN(x1, x2, x3)  =  APP_IN(x2)

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

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:

suffix_in(Xs, Ys) → U1(Xs, Ys, app_in(X, Xs, Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U2(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U1(Xs, Ys, app_out(X, Xs, Ys)) → suffix_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
suffix_in(x1, x2)  =  suffix_in(x1)
U1(x1, x2, x3)  =  U1(x1, x3)
app_in(x1, x2, x3)  =  app_in(x2)
.(x1, x2)  =  .(x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x5)
[]  =  []
app_out(x1, x2, x3)  =  app_out(x1, x2, x3)
suffix_out(x1, x2)  =  suffix_out(x1, x2)
APP_IN(x1, x2, x3)  =  APP_IN(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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)  =  .(x2)
APP_IN(x1, x2, x3)  =  APP_IN(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ NonTerminationProof

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

APP_IN(Ys) → APP_IN(Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_IN(Ys) → APP_IN(Ys)

The TRS R consists of the following rules:none


s = APP_IN(Ys) evaluates to t =APP_IN(Ys)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_IN(Ys) to APP_IN(Ys).