Left Termination of the query pattern p_in_1(g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

p(0).
p(s(X)) :- ','(geq(X, Y), p(Y)).
geq(X, X).
geq(s(X), Y) :- geq(X, Y).

Queries:

p(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:

p_in(s(X)) → U1(X, geq_in(X, Y))
geq_in(s(X), Y) → U3(X, Y, geq_in(X, Y))
geq_in(X, X) → geq_out(X, X)
U3(X, Y, geq_out(X, Y)) → geq_out(s(X), Y)
U1(X, geq_out(X, Y)) → U2(X, p_in(Y))
p_in(0) → p_out(0)
U2(X, p_out(Y)) → p_out(s(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
geq_in(x1, x2)  =  geq_in(x1)
U3(x1, x2, x3)  =  U3(x3)
geq_out(x1, x2)  =  geq_out(x2)
U2(x1, x2)  =  U2(x2)
0  =  0
p_out(x1)  =  p_out

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:

p_in(s(X)) → U1(X, geq_in(X, Y))
geq_in(s(X), Y) → U3(X, Y, geq_in(X, Y))
geq_in(X, X) → geq_out(X, X)
U3(X, Y, geq_out(X, Y)) → geq_out(s(X), Y)
U1(X, geq_out(X, Y)) → U2(X, p_in(Y))
p_in(0) → p_out(0)
U2(X, p_out(Y)) → p_out(s(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
geq_in(x1, x2)  =  geq_in(x1)
U3(x1, x2, x3)  =  U3(x3)
geq_out(x1, x2)  =  geq_out(x2)
U2(x1, x2)  =  U2(x2)
0  =  0
p_out(x1)  =  p_out


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

P_IN(s(X)) → U11(X, geq_in(X, Y))
P_IN(s(X)) → GEQ_IN(X, Y)
GEQ_IN(s(X), Y) → U31(X, Y, geq_in(X, Y))
GEQ_IN(s(X), Y) → GEQ_IN(X, Y)
U11(X, geq_out(X, Y)) → U21(X, p_in(Y))
U11(X, geq_out(X, Y)) → P_IN(Y)

The TRS R consists of the following rules:

p_in(s(X)) → U1(X, geq_in(X, Y))
geq_in(s(X), Y) → U3(X, Y, geq_in(X, Y))
geq_in(X, X) → geq_out(X, X)
U3(X, Y, geq_out(X, Y)) → geq_out(s(X), Y)
U1(X, geq_out(X, Y)) → U2(X, p_in(Y))
p_in(0) → p_out(0)
U2(X, p_out(Y)) → p_out(s(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
geq_in(x1, x2)  =  geq_in(x1)
U3(x1, x2, x3)  =  U3(x3)
geq_out(x1, x2)  =  geq_out(x2)
U2(x1, x2)  =  U2(x2)
0  =  0
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN(x1)
U31(x1, x2, x3)  =  U31(x3)
GEQ_IN(x1, x2)  =  GEQ_IN(x1)
U11(x1, x2)  =  U11(x2)
U21(x1, x2)  =  U21(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:

P_IN(s(X)) → U11(X, geq_in(X, Y))
P_IN(s(X)) → GEQ_IN(X, Y)
GEQ_IN(s(X), Y) → U31(X, Y, geq_in(X, Y))
GEQ_IN(s(X), Y) → GEQ_IN(X, Y)
U11(X, geq_out(X, Y)) → U21(X, p_in(Y))
U11(X, geq_out(X, Y)) → P_IN(Y)

The TRS R consists of the following rules:

p_in(s(X)) → U1(X, geq_in(X, Y))
geq_in(s(X), Y) → U3(X, Y, geq_in(X, Y))
geq_in(X, X) → geq_out(X, X)
U3(X, Y, geq_out(X, Y)) → geq_out(s(X), Y)
U1(X, geq_out(X, Y)) → U2(X, p_in(Y))
p_in(0) → p_out(0)
U2(X, p_out(Y)) → p_out(s(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
geq_in(x1, x2)  =  geq_in(x1)
U3(x1, x2, x3)  =  U3(x3)
geq_out(x1, x2)  =  geq_out(x2)
U2(x1, x2)  =  U2(x2)
0  =  0
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN(x1)
U31(x1, x2, x3)  =  U31(x3)
GEQ_IN(x1, x2)  =  GEQ_IN(x1)
U11(x1, x2)  =  U11(x2)
U21(x1, x2)  =  U21(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

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

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

GEQ_IN(s(X), Y) → GEQ_IN(X, Y)

The TRS R consists of the following rules:

p_in(s(X)) → U1(X, geq_in(X, Y))
geq_in(s(X), Y) → U3(X, Y, geq_in(X, Y))
geq_in(X, X) → geq_out(X, X)
U3(X, Y, geq_out(X, Y)) → geq_out(s(X), Y)
U1(X, geq_out(X, Y)) → U2(X, p_in(Y))
p_in(0) → p_out(0)
U2(X, p_out(Y)) → p_out(s(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
geq_in(x1, x2)  =  geq_in(x1)
U3(x1, x2, x3)  =  U3(x3)
geq_out(x1, x2)  =  geq_out(x2)
U2(x1, x2)  =  U2(x2)
0  =  0
p_out(x1)  =  p_out
GEQ_IN(x1, x2)  =  GEQ_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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

GEQ_IN(s(X), Y) → GEQ_IN(X, Y)

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

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

GEQ_IN(s(X)) → GEQ_IN(X)

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:

P_IN(s(X)) → U11(X, geq_in(X, Y))
U11(X, geq_out(X, Y)) → P_IN(Y)

The TRS R consists of the following rules:

p_in(s(X)) → U1(X, geq_in(X, Y))
geq_in(s(X), Y) → U3(X, Y, geq_in(X, Y))
geq_in(X, X) → geq_out(X, X)
U3(X, Y, geq_out(X, Y)) → geq_out(s(X), Y)
U1(X, geq_out(X, Y)) → U2(X, p_in(Y))
p_in(0) → p_out(0)
U2(X, p_out(Y)) → p_out(s(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
geq_in(x1, x2)  =  geq_in(x1)
U3(x1, x2, x3)  =  U3(x3)
geq_out(x1, x2)  =  geq_out(x2)
U2(x1, x2)  =  U2(x2)
0  =  0
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN(x1)
U11(x1, x2)  =  U11(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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

P_IN(s(X)) → U11(X, geq_in(X, Y))
U11(X, geq_out(X, Y)) → P_IN(Y)

The TRS R consists of the following rules:

geq_in(s(X), Y) → U3(X, Y, geq_in(X, Y))
geq_in(X, X) → geq_out(X, X)
U3(X, Y, geq_out(X, Y)) → geq_out(s(X), Y)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
geq_in(x1, x2)  =  geq_in(x1)
U3(x1, x2, x3)  =  U3(x3)
geq_out(x1, x2)  =  geq_out(x2)
P_IN(x1)  =  P_IN(x1)
U11(x1, x2)  =  U11(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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof

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

U11(geq_out(Y)) → P_IN(Y)
P_IN(s(X)) → U11(geq_in(X))

The TRS R consists of the following rules:

geq_in(s(X)) → U3(geq_in(X))
geq_in(X) → geq_out(X)
U3(geq_out(Y)) → geq_out(Y)

The set Q consists of the following terms:

geq_in(x0)
U3(x0)

We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

U11(geq_out(Y)) → P_IN(Y)
P_IN(s(X)) → U11(geq_in(X))
The following rules are removed from R:

geq_in(s(X)) → U3(geq_in(X))
U3(geq_out(Y)) → geq_out(Y)
Used ordering: POLO with Polynomial interpretation [25]:

POL(P_IN(x1)) = 2 + x1   
POL(U11(x1)) = 1 + x1   
POL(U3(x1)) = 2·x1   
POL(geq_in(x1)) = 2 + x1   
POL(geq_out(x1)) = 2 + x1   
POL(s(x1)) = 2 + 2·x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof
                          ↳ QDP
                            ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

geq_in(X) → geq_out(X)

The set Q consists of the following terms:

geq_in(x0)
U3(x0)

We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.