Left Termination of the query pattern even_in_1(g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

even(0).
even(s(s(0))).
even(s(s(s(X)))) :- odd(X).
odd(s(0)).
odd(s(X)) :- even(s(s(X))).

Queries:

even(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:

even_in(s(s(s(X)))) → U1(X, odd_in(X))
odd_in(s(X)) → U2(X, even_in(s(s(X))))
even_in(s(s(0))) → even_out(s(s(0)))
even_in(0) → even_out(0)
U2(X, even_out(s(s(X)))) → odd_out(s(X))
odd_in(s(0)) → odd_out(s(0))
U1(X, odd_out(X)) → even_out(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in(x1)  =  even_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
odd_in(x1)  =  odd_in(x1)
U2(x1, x2)  =  U2(x2)
0  =  0
even_out(x1)  =  even_out
odd_out(x1)  =  odd_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:

even_in(s(s(s(X)))) → U1(X, odd_in(X))
odd_in(s(X)) → U2(X, even_in(s(s(X))))
even_in(s(s(0))) → even_out(s(s(0)))
even_in(0) → even_out(0)
U2(X, even_out(s(s(X)))) → odd_out(s(X))
odd_in(s(0)) → odd_out(s(0))
U1(X, odd_out(X)) → even_out(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in(x1)  =  even_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
odd_in(x1)  =  odd_in(x1)
U2(x1, x2)  =  U2(x2)
0  =  0
even_out(x1)  =  even_out
odd_out(x1)  =  odd_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:

EVEN_IN(s(s(s(X)))) → U11(X, odd_in(X))
EVEN_IN(s(s(s(X)))) → ODD_IN(X)
ODD_IN(s(X)) → U21(X, even_in(s(s(X))))
ODD_IN(s(X)) → EVEN_IN(s(s(X)))

The TRS R consists of the following rules:

even_in(s(s(s(X)))) → U1(X, odd_in(X))
odd_in(s(X)) → U2(X, even_in(s(s(X))))
even_in(s(s(0))) → even_out(s(s(0)))
even_in(0) → even_out(0)
U2(X, even_out(s(s(X)))) → odd_out(s(X))
odd_in(s(0)) → odd_out(s(0))
U1(X, odd_out(X)) → even_out(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in(x1)  =  even_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
odd_in(x1)  =  odd_in(x1)
U2(x1, x2)  =  U2(x2)
0  =  0
even_out(x1)  =  even_out
odd_out(x1)  =  odd_out
EVEN_IN(x1)  =  EVEN_IN(x1)
ODD_IN(x1)  =  ODD_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:

EVEN_IN(s(s(s(X)))) → U11(X, odd_in(X))
EVEN_IN(s(s(s(X)))) → ODD_IN(X)
ODD_IN(s(X)) → U21(X, even_in(s(s(X))))
ODD_IN(s(X)) → EVEN_IN(s(s(X)))

The TRS R consists of the following rules:

even_in(s(s(s(X)))) → U1(X, odd_in(X))
odd_in(s(X)) → U2(X, even_in(s(s(X))))
even_in(s(s(0))) → even_out(s(s(0)))
even_in(0) → even_out(0)
U2(X, even_out(s(s(X)))) → odd_out(s(X))
odd_in(s(0)) → odd_out(s(0))
U1(X, odd_out(X)) → even_out(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in(x1)  =  even_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
odd_in(x1)  =  odd_in(x1)
U2(x1, x2)  =  U2(x2)
0  =  0
even_out(x1)  =  even_out
odd_out(x1)  =  odd_out
EVEN_IN(x1)  =  EVEN_IN(x1)
ODD_IN(x1)  =  ODD_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 1 SCC with 2 less nodes.

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

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

ODD_IN(s(X)) → EVEN_IN(s(s(X)))
EVEN_IN(s(s(s(X)))) → ODD_IN(X)

The TRS R consists of the following rules:

even_in(s(s(s(X)))) → U1(X, odd_in(X))
odd_in(s(X)) → U2(X, even_in(s(s(X))))
even_in(s(s(0))) → even_out(s(s(0)))
even_in(0) → even_out(0)
U2(X, even_out(s(s(X)))) → odd_out(s(X))
odd_in(s(0)) → odd_out(s(0))
U1(X, odd_out(X)) → even_out(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in(x1)  =  even_in(x1)
s(x1)  =  s(x1)
U1(x1, x2)  =  U1(x2)
odd_in(x1)  =  odd_in(x1)
U2(x1, x2)  =  U2(x2)
0  =  0
even_out(x1)  =  even_out
odd_out(x1)  =  odd_out
EVEN_IN(x1)  =  EVEN_IN(x1)
ODD_IN(x1)  =  ODD_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

ODD_IN(s(X)) → EVEN_IN(s(s(X)))
EVEN_IN(s(s(s(X)))) → ODD_IN(X)

R is empty.
Pi is empty.
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
                      ↳ RuleRemovalProof

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

ODD_IN(s(X)) → EVEN_IN(s(s(X)))
EVEN_IN(s(s(s(X)))) → ODD_IN(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

EVEN_IN(s(s(s(X)))) → ODD_IN(X)


Used ordering: POLO with Polynomial interpretation [25]:

POL(EVEN_IN(x1)) = x1   
POL(ODD_IN(x1)) = 2 + 2·x1   
POL(s(x1)) = 2 + 2·x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ DependencyGraphProof

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

ODD_IN(s(X)) → EVEN_IN(s(s(X)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.