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].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(X, s(s(s(s(0))))))
lte_in(0, Y) → lte_out(0, Y)
lte_in(s(X), s(Y)) → U2(X, Y, lte_in(X, Y))
U2(X, Y, lte_out(X, Y)) → lte_out(s(X), s(Y))
U3(lte_out(X, s(s(s(s(0)))))) → U4(even_in(X))
even_in(0) → even_out(0)
even_in(s(s(X))) → U1(X, even_in(X))
U1(X, even_out(X)) → even_out(s(s(X)))
U4(even_out(X)) → goal_out

The argument filtering Pi contains the following mapping:
goal_in  =  goal_in
U3(x1)  =  U3(x1)
lte_in(x1, x2)  =  lte_in(x2)
s(x1)  =  s(x1)
0  =  0
lte_out(x1, x2)  =  lte_out(x1)
U2(x1, x2, x3)  =  U2(x3)
U4(x1)  =  U4(x1)
even_in(x1)  =  even_in(x1)
even_out(x1)  =  even_out
U1(x1, x2)  =  U1(x2)
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(X, s(s(s(s(0))))))
lte_in(0, Y) → lte_out(0, Y)
lte_in(s(X), s(Y)) → U2(X, Y, lte_in(X, Y))
U2(X, Y, lte_out(X, Y)) → lte_out(s(X), s(Y))
U3(lte_out(X, s(s(s(s(0)))))) → U4(even_in(X))
even_in(0) → even_out(0)
even_in(s(s(X))) → U1(X, even_in(X))
U1(X, even_out(X)) → even_out(s(s(X)))
U4(even_out(X)) → goal_out

The argument filtering Pi contains the following mapping:
goal_in  =  goal_in
U3(x1)  =  U3(x1)
lte_in(x1, x2)  =  lte_in(x2)
s(x1)  =  s(x1)
0  =  0
lte_out(x1, x2)  =  lte_out(x1)
U2(x1, x2, x3)  =  U2(x3)
U4(x1)  =  U4(x1)
even_in(x1)  =  even_in(x1)
even_out(x1)  =  even_out
U1(x1, x2)  =  U1(x2)
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 → U31(lte_in(X, s(s(s(s(0))))))
GOAL_IN → LTE_IN(X, s(s(s(s(0)))))
LTE_IN(s(X), s(Y)) → U21(X, Y, lte_in(X, Y))
LTE_IN(s(X), s(Y)) → LTE_IN(X, Y)
U31(lte_out(X, s(s(s(s(0)))))) → U41(even_in(X))
U31(lte_out(X, s(s(s(s(0)))))) → EVEN_IN(X)
EVEN_IN(s(s(X))) → U11(X, even_in(X))
EVEN_IN(s(s(X))) → EVEN_IN(X)

The TRS R consists of the following rules:

goal_in → U3(lte_in(X, s(s(s(s(0))))))
lte_in(0, Y) → lte_out(0, Y)
lte_in(s(X), s(Y)) → U2(X, Y, lte_in(X, Y))
U2(X, Y, lte_out(X, Y)) → lte_out(s(X), s(Y))
U3(lte_out(X, s(s(s(s(0)))))) → U4(even_in(X))
even_in(0) → even_out(0)
even_in(s(s(X))) → U1(X, even_in(X))
U1(X, even_out(X)) → even_out(s(s(X)))
U4(even_out(X)) → goal_out

The argument filtering Pi contains the following mapping:
goal_in  =  goal_in
U3(x1)  =  U3(x1)
lte_in(x1, x2)  =  lte_in(x2)
s(x1)  =  s(x1)
0  =  0
lte_out(x1, x2)  =  lte_out(x1)
U2(x1, x2, x3)  =  U2(x3)
U4(x1)  =  U4(x1)
even_in(x1)  =  even_in(x1)
even_out(x1)  =  even_out
U1(x1, x2)  =  U1(x2)
goal_out  =  goal_out
EVEN_IN(x1)  =  EVEN_IN(x1)
U41(x1)  =  U41(x1)
U31(x1)  =  U31(x1)
U11(x1, x2)  =  U11(x2)
LTE_IN(x1, x2)  =  LTE_IN(x2)
GOAL_IN  =  GOAL_IN
U21(x1, x2, x3)  =  U21(x3)

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 → U31(lte_in(X, s(s(s(s(0))))))
GOAL_IN → LTE_IN(X, s(s(s(s(0)))))
LTE_IN(s(X), s(Y)) → U21(X, Y, lte_in(X, Y))
LTE_IN(s(X), s(Y)) → LTE_IN(X, Y)
U31(lte_out(X, s(s(s(s(0)))))) → U41(even_in(X))
U31(lte_out(X, s(s(s(s(0)))))) → EVEN_IN(X)
EVEN_IN(s(s(X))) → U11(X, even_in(X))
EVEN_IN(s(s(X))) → EVEN_IN(X)

The TRS R consists of the following rules:

goal_in → U3(lte_in(X, s(s(s(s(0))))))
lte_in(0, Y) → lte_out(0, Y)
lte_in(s(X), s(Y)) → U2(X, Y, lte_in(X, Y))
U2(X, Y, lte_out(X, Y)) → lte_out(s(X), s(Y))
U3(lte_out(X, s(s(s(s(0)))))) → U4(even_in(X))
even_in(0) → even_out(0)
even_in(s(s(X))) → U1(X, even_in(X))
U1(X, even_out(X)) → even_out(s(s(X)))
U4(even_out(X)) → goal_out

The argument filtering Pi contains the following mapping:
goal_in  =  goal_in
U3(x1)  =  U3(x1)
lte_in(x1, x2)  =  lte_in(x2)
s(x1)  =  s(x1)
0  =  0
lte_out(x1, x2)  =  lte_out(x1)
U2(x1, x2, x3)  =  U2(x3)
U4(x1)  =  U4(x1)
even_in(x1)  =  even_in(x1)
even_out(x1)  =  even_out
U1(x1, x2)  =  U1(x2)
goal_out  =  goal_out
EVEN_IN(x1)  =  EVEN_IN(x1)
U41(x1)  =  U41(x1)
U31(x1)  =  U31(x1)
U11(x1, x2)  =  U11(x2)
LTE_IN(x1, x2)  =  LTE_IN(x2)
GOAL_IN  =  GOAL_IN
U21(x1, x2, x3)  =  U21(x3)

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(s(s(X))) → EVEN_IN(X)

The TRS R consists of the following rules:

goal_in → U3(lte_in(X, s(s(s(s(0))))))
lte_in(0, Y) → lte_out(0, Y)
lte_in(s(X), s(Y)) → U2(X, Y, lte_in(X, Y))
U2(X, Y, lte_out(X, Y)) → lte_out(s(X), s(Y))
U3(lte_out(X, s(s(s(s(0)))))) → U4(even_in(X))
even_in(0) → even_out(0)
even_in(s(s(X))) → U1(X, even_in(X))
U1(X, even_out(X)) → even_out(s(s(X)))
U4(even_out(X)) → goal_out

The argument filtering Pi contains the following mapping:
goal_in  =  goal_in
U3(x1)  =  U3(x1)
lte_in(x1, x2)  =  lte_in(x2)
s(x1)  =  s(x1)
0  =  0
lte_out(x1, x2)  =  lte_out(x1)
U2(x1, x2, x3)  =  U2(x3)
U4(x1)  =  U4(x1)
even_in(x1)  =  even_in(x1)
even_out(x1)  =  even_out
U1(x1, x2)  =  U1(x2)
goal_out  =  goal_out
EVEN_IN(x1)  =  EVEN_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:

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

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

EVEN_IN(s(s(X))) → EVEN_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:

LTE_IN(s(X), s(Y)) → LTE_IN(X, Y)

The TRS R consists of the following rules:

goal_in → U3(lte_in(X, s(s(s(s(0))))))
lte_in(0, Y) → lte_out(0, Y)
lte_in(s(X), s(Y)) → U2(X, Y, lte_in(X, Y))
U2(X, Y, lte_out(X, Y)) → lte_out(s(X), s(Y))
U3(lte_out(X, s(s(s(s(0)))))) → U4(even_in(X))
even_in(0) → even_out(0)
even_in(s(s(X))) → U1(X, even_in(X))
U1(X, even_out(X)) → even_out(s(s(X)))
U4(even_out(X)) → goal_out

The argument filtering Pi contains the following mapping:
goal_in  =  goal_in
U3(x1)  =  U3(x1)
lte_in(x1, x2)  =  lte_in(x2)
s(x1)  =  s(x1)
0  =  0
lte_out(x1, x2)  =  lte_out(x1)
U2(x1, x2, x3)  =  U2(x3)
U4(x1)  =  U4(x1)
even_in(x1)  =  even_in(x1)
even_out(x1)  =  even_out
U1(x1, x2)  =  U1(x2)
goal_out  =  goal_out
LTE_IN(x1, x2)  =  LTE_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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

LTE_IN(s(X), s(Y)) → LTE_IN(X, Y)

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

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

LTE_IN(s(Y)) → LTE_IN(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: