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



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

even(s(s(X))) :- even(X).
even(0).
lte(s(X), s(Y)) :- lte(X, Y).
lte(0, Y).
goal :- ','(lte(X, s(s(s(s(0))))), even(X)).

Queries:

goal().

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
lte_in: (f,b)
even_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal_in_ → U3_(lte_in_ag(X, s(s(s(s(0))))))
lte_in_ag(s(X), s(Y)) → U2_ag(X, Y, lte_in_ag(X, Y))
lte_in_ag(0, Y) → lte_out_ag(0, Y)
U2_ag(X, Y, lte_out_ag(X, Y)) → lte_out_ag(s(X), s(Y))
U3_(lte_out_ag(X, s(s(s(s(0)))))) → U4_(even_in_g(X))
even_in_g(s(s(X))) → U1_g(X, even_in_g(X))
even_in_g(0) → even_out_g(0)
U1_g(X, even_out_g(X)) → even_out_g(s(s(X)))
U4_(even_out_g(X)) → goal_out_

The argument filtering Pi contains the following mapping:
goal_in_  =  goal_in_
U3_(x1)  =  U3_(x1)
lte_in_ag(x1, x2)  =  lte_in_ag(x2)
s(x1)  =  s(x1)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
lte_out_ag(x1, x2)  =  lte_out_ag(x1)
0  =  0
U4_(x1)  =  U4_(x1)
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
even_out_g(x1)  =  even_out_g
goal_out_  =  goal_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:

goal_in_ → U3_(lte_in_ag(X, s(s(s(s(0))))))
lte_in_ag(s(X), s(Y)) → U2_ag(X, Y, lte_in_ag(X, Y))
lte_in_ag(0, Y) → lte_out_ag(0, Y)
U2_ag(X, Y, lte_out_ag(X, Y)) → lte_out_ag(s(X), s(Y))
U3_(lte_out_ag(X, s(s(s(s(0)))))) → U4_(even_in_g(X))
even_in_g(s(s(X))) → U1_g(X, even_in_g(X))
even_in_g(0) → even_out_g(0)
U1_g(X, even_out_g(X)) → even_out_g(s(s(X)))
U4_(even_out_g(X)) → goal_out_

The argument filtering Pi contains the following mapping:
goal_in_  =  goal_in_
U3_(x1)  =  U3_(x1)
lte_in_ag(x1, x2)  =  lte_in_ag(x2)
s(x1)  =  s(x1)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
lte_out_ag(x1, x2)  =  lte_out_ag(x1)
0  =  0
U4_(x1)  =  U4_(x1)
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
even_out_g(x1)  =  even_out_g
goal_out_  =  goal_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:

GOAL_IN_ → U3_1(lte_in_ag(X, s(s(s(s(0))))))
GOAL_IN_ → LTE_IN_AG(X, s(s(s(s(0)))))
LTE_IN_AG(s(X), s(Y)) → U2_AG(X, Y, lte_in_ag(X, Y))
LTE_IN_AG(s(X), s(Y)) → LTE_IN_AG(X, Y)
U3_1(lte_out_ag(X, s(s(s(s(0)))))) → U4_1(even_in_g(X))
U3_1(lte_out_ag(X, s(s(s(s(0)))))) → EVEN_IN_G(X)
EVEN_IN_G(s(s(X))) → U1_G(X, even_in_g(X))
EVEN_IN_G(s(s(X))) → EVEN_IN_G(X)

The TRS R consists of the following rules:

goal_in_ → U3_(lte_in_ag(X, s(s(s(s(0))))))
lte_in_ag(s(X), s(Y)) → U2_ag(X, Y, lte_in_ag(X, Y))
lte_in_ag(0, Y) → lte_out_ag(0, Y)
U2_ag(X, Y, lte_out_ag(X, Y)) → lte_out_ag(s(X), s(Y))
U3_(lte_out_ag(X, s(s(s(s(0)))))) → U4_(even_in_g(X))
even_in_g(s(s(X))) → U1_g(X, even_in_g(X))
even_in_g(0) → even_out_g(0)
U1_g(X, even_out_g(X)) → even_out_g(s(s(X)))
U4_(even_out_g(X)) → goal_out_

The argument filtering Pi contains the following mapping:
goal_in_  =  goal_in_
U3_(x1)  =  U3_(x1)
lte_in_ag(x1, x2)  =  lte_in_ag(x2)
s(x1)  =  s(x1)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
lte_out_ag(x1, x2)  =  lte_out_ag(x1)
0  =  0
U4_(x1)  =  U4_(x1)
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
even_out_g(x1)  =  even_out_g
goal_out_  =  goal_out_
U3_1(x1)  =  U3_1(x1)
U4_1(x1)  =  U4_1(x1)
GOAL_IN_  =  GOAL_IN_
U1_G(x1, x2)  =  U1_G(x2)
U2_AG(x1, x2, x3)  =  U2_AG(x3)
LTE_IN_AG(x1, x2)  =  LTE_IN_AG(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:

GOAL_IN_ → U3_1(lte_in_ag(X, s(s(s(s(0))))))
GOAL_IN_ → LTE_IN_AG(X, s(s(s(s(0)))))
LTE_IN_AG(s(X), s(Y)) → U2_AG(X, Y, lte_in_ag(X, Y))
LTE_IN_AG(s(X), s(Y)) → LTE_IN_AG(X, Y)
U3_1(lte_out_ag(X, s(s(s(s(0)))))) → U4_1(even_in_g(X))
U3_1(lte_out_ag(X, s(s(s(s(0)))))) → EVEN_IN_G(X)
EVEN_IN_G(s(s(X))) → U1_G(X, even_in_g(X))
EVEN_IN_G(s(s(X))) → EVEN_IN_G(X)

The TRS R consists of the following rules:

goal_in_ → U3_(lte_in_ag(X, s(s(s(s(0))))))
lte_in_ag(s(X), s(Y)) → U2_ag(X, Y, lte_in_ag(X, Y))
lte_in_ag(0, Y) → lte_out_ag(0, Y)
U2_ag(X, Y, lte_out_ag(X, Y)) → lte_out_ag(s(X), s(Y))
U3_(lte_out_ag(X, s(s(s(s(0)))))) → U4_(even_in_g(X))
even_in_g(s(s(X))) → U1_g(X, even_in_g(X))
even_in_g(0) → even_out_g(0)
U1_g(X, even_out_g(X)) → even_out_g(s(s(X)))
U4_(even_out_g(X)) → goal_out_

The argument filtering Pi contains the following mapping:
goal_in_  =  goal_in_
U3_(x1)  =  U3_(x1)
lte_in_ag(x1, x2)  =  lte_in_ag(x2)
s(x1)  =  s(x1)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
lte_out_ag(x1, x2)  =  lte_out_ag(x1)
0  =  0
U4_(x1)  =  U4_(x1)
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
even_out_g(x1)  =  even_out_g
goal_out_  =  goal_out_
U3_1(x1)  =  U3_1(x1)
U4_1(x1)  =  U4_1(x1)
GOAL_IN_  =  GOAL_IN_
U1_G(x1, x2)  =  U1_G(x2)
U2_AG(x1, x2, x3)  =  U2_AG(x3)
LTE_IN_AG(x1, x2)  =  LTE_IN_AG(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 2 SCCs with 6 less nodes.

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

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

EVEN_IN_G(s(s(X))) → EVEN_IN_G(X)

The TRS R consists of the following rules:

goal_in_ → U3_(lte_in_ag(X, s(s(s(s(0))))))
lte_in_ag(s(X), s(Y)) → U2_ag(X, Y, lte_in_ag(X, Y))
lte_in_ag(0, Y) → lte_out_ag(0, Y)
U2_ag(X, Y, lte_out_ag(X, Y)) → lte_out_ag(s(X), s(Y))
U3_(lte_out_ag(X, s(s(s(s(0)))))) → U4_(even_in_g(X))
even_in_g(s(s(X))) → U1_g(X, even_in_g(X))
even_in_g(0) → even_out_g(0)
U1_g(X, even_out_g(X)) → even_out_g(s(s(X)))
U4_(even_out_g(X)) → goal_out_

The argument filtering Pi contains the following mapping:
goal_in_  =  goal_in_
U3_(x1)  =  U3_(x1)
lte_in_ag(x1, x2)  =  lte_in_ag(x2)
s(x1)  =  s(x1)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
lte_out_ag(x1, x2)  =  lte_out_ag(x1)
0  =  0
U4_(x1)  =  U4_(x1)
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
even_out_g(x1)  =  even_out_g
goal_out_  =  goal_out_
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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

EVEN_IN_G(s(s(X))) → EVEN_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

EVEN_IN_G(s(s(X))) → EVEN_IN_G(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:

LTE_IN_AG(s(X), s(Y)) → LTE_IN_AG(X, Y)

The TRS R consists of the following rules:

goal_in_ → U3_(lte_in_ag(X, s(s(s(s(0))))))
lte_in_ag(s(X), s(Y)) → U2_ag(X, Y, lte_in_ag(X, Y))
lte_in_ag(0, Y) → lte_out_ag(0, Y)
U2_ag(X, Y, lte_out_ag(X, Y)) → lte_out_ag(s(X), s(Y))
U3_(lte_out_ag(X, s(s(s(s(0)))))) → U4_(even_in_g(X))
even_in_g(s(s(X))) → U1_g(X, even_in_g(X))
even_in_g(0) → even_out_g(0)
U1_g(X, even_out_g(X)) → even_out_g(s(s(X)))
U4_(even_out_g(X)) → goal_out_

The argument filtering Pi contains the following mapping:
goal_in_  =  goal_in_
U3_(x1)  =  U3_(x1)
lte_in_ag(x1, x2)  =  lte_in_ag(x2)
s(x1)  =  s(x1)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
lte_out_ag(x1, x2)  =  lte_out_ag(x1)
0  =  0
U4_(x1)  =  U4_(x1)
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
even_out_g(x1)  =  even_out_g
goal_out_  =  goal_out_
LTE_IN_AG(x1, x2)  =  LTE_IN_AG(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:

LTE_IN_AG(s(X), s(Y)) → LTE_IN_AG(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LTE_IN_AG(x1, x2)  =  LTE_IN_AG(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
                        ↳ QDPSizeChangeProof

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

LTE_IN_AG(s(Y)) → LTE_IN_AG(Y)

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: