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]. With regard to the inferred argument filtering the predicates were used in the following modes:
even_in: (b)
odd_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

even_in_g(0) → even_out_g(0)
even_in_g(s(s(0))) → even_out_g(s(s(0)))
even_in_g(s(s(s(X)))) → U1_g(X, odd_in_g(X))
odd_in_g(s(0)) → odd_out_g(s(0))
odd_in_g(s(X)) → U2_g(X, even_in_g(s(s(X))))
U2_g(X, even_out_g(s(s(X)))) → odd_out_g(s(X))
U1_g(X, odd_out_g(X)) → even_out_g(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
0  =  0
even_out_g(x1)  =  even_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
odd_in_g(x1)  =  odd_in_g(x1)
odd_out_g(x1)  =  odd_out_g
U2_g(x1, x2)  =  U2_g(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:

even_in_g(0) → even_out_g(0)
even_in_g(s(s(0))) → even_out_g(s(s(0)))
even_in_g(s(s(s(X)))) → U1_g(X, odd_in_g(X))
odd_in_g(s(0)) → odd_out_g(s(0))
odd_in_g(s(X)) → U2_g(X, even_in_g(s(s(X))))
U2_g(X, even_out_g(s(s(X)))) → odd_out_g(s(X))
U1_g(X, odd_out_g(X)) → even_out_g(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
0  =  0
even_out_g(x1)  =  even_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
odd_in_g(x1)  =  odd_in_g(x1)
odd_out_g(x1)  =  odd_out_g
U2_g(x1, x2)  =  U2_g(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:

EVEN_IN_G(s(s(s(X)))) → U1_G(X, odd_in_g(X))
EVEN_IN_G(s(s(s(X)))) → ODD_IN_G(X)
ODD_IN_G(s(X)) → U2_G(X, even_in_g(s(s(X))))
ODD_IN_G(s(X)) → EVEN_IN_G(s(s(X)))

The TRS R consists of the following rules:

even_in_g(0) → even_out_g(0)
even_in_g(s(s(0))) → even_out_g(s(s(0)))
even_in_g(s(s(s(X)))) → U1_g(X, odd_in_g(X))
odd_in_g(s(0)) → odd_out_g(s(0))
odd_in_g(s(X)) → U2_g(X, even_in_g(s(s(X))))
U2_g(X, even_out_g(s(s(X)))) → odd_out_g(s(X))
U1_g(X, odd_out_g(X)) → even_out_g(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
0  =  0
even_out_g(x1)  =  even_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
odd_in_g(x1)  =  odd_in_g(x1)
odd_out_g(x1)  =  odd_out_g
U2_g(x1, x2)  =  U2_g(x2)
U2_G(x1, x2)  =  U2_G(x2)
ODD_IN_G(x1)  =  ODD_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
EVEN_IN_G(x1)  =  EVEN_IN_G(x1)

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_G(s(s(s(X)))) → U1_G(X, odd_in_g(X))
EVEN_IN_G(s(s(s(X)))) → ODD_IN_G(X)
ODD_IN_G(s(X)) → U2_G(X, even_in_g(s(s(X))))
ODD_IN_G(s(X)) → EVEN_IN_G(s(s(X)))

The TRS R consists of the following rules:

even_in_g(0) → even_out_g(0)
even_in_g(s(s(0))) → even_out_g(s(s(0)))
even_in_g(s(s(s(X)))) → U1_g(X, odd_in_g(X))
odd_in_g(s(0)) → odd_out_g(s(0))
odd_in_g(s(X)) → U2_g(X, even_in_g(s(s(X))))
U2_g(X, even_out_g(s(s(X)))) → odd_out_g(s(X))
U1_g(X, odd_out_g(X)) → even_out_g(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
0  =  0
even_out_g(x1)  =  even_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
odd_in_g(x1)  =  odd_in_g(x1)
odd_out_g(x1)  =  odd_out_g
U2_g(x1, x2)  =  U2_g(x2)
U2_G(x1, x2)  =  U2_G(x2)
ODD_IN_G(x1)  =  ODD_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
EVEN_IN_G(x1)  =  EVEN_IN_G(x1)

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_G(s(X)) → EVEN_IN_G(s(s(X)))
EVEN_IN_G(s(s(s(X)))) → ODD_IN_G(X)

The TRS R consists of the following rules:

even_in_g(0) → even_out_g(0)
even_in_g(s(s(0))) → even_out_g(s(s(0)))
even_in_g(s(s(s(X)))) → U1_g(X, odd_in_g(X))
odd_in_g(s(0)) → odd_out_g(s(0))
odd_in_g(s(X)) → U2_g(X, even_in_g(s(s(X))))
U2_g(X, even_out_g(s(s(X)))) → odd_out_g(s(X))
U1_g(X, odd_out_g(X)) → even_out_g(s(s(s(X))))

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
0  =  0
even_out_g(x1)  =  even_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
odd_in_g(x1)  =  odd_in_g(x1)
odd_out_g(x1)  =  odd_out_g
U2_g(x1, x2)  =  U2_g(x2)
ODD_IN_G(x1)  =  ODD_IN_G(x1)
EVEN_IN_G(x1)  =  EVEN_IN_G(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_G(s(X)) → EVEN_IN_G(s(s(X)))
EVEN_IN_G(s(s(s(X)))) → ODD_IN_G(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_G(s(X)) → EVEN_IN_G(s(s(X)))
EVEN_IN_G(s(s(s(X)))) → ODD_IN_G(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_G(s(s(s(X)))) → ODD_IN_G(X)


Used ordering: POLO with Polynomial interpretation [25]:

POL(EVEN_IN_G(x1)) = x1   
POL(ODD_IN_G(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_G(s(X)) → EVEN_IN_G(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.