Left Termination of the query pattern in_in_2(g, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

in(X, tree(X, X1, X2)).
in(X, tree(Y, Left, X1)) :- ','(less(X, Y), in(X, Left)).
in(X, tree(Y, X1, Right)) :- ','(less(Y, X), in(X, Right)).
less(0, s(X)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

in(g,a).

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:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
in_out(x1, x2)  =  in_out

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
in_out(x1, x2)  =  in_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:

IN_IN(X, tree(Y, X1, Right)) → U31(X, Y, X1, Right, less_in(Y, X))
IN_IN(X, tree(Y, X1, Right)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U51(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U31(X, Y, X1, Right, less_out(Y, X)) → U41(X, Y, X1, Right, in_in(X, Right))
U31(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
IN_IN(X, tree(Y, Left, X1)) → U11(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, Left, X1)) → LESS_IN(X, Y)
U11(X, Y, Left, X1, less_out(X, Y)) → U21(X, Y, Left, X1, in_in(X, Left))
U11(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
in_out(x1, x2)  =  in_out
IN_IN(x1, x2)  =  IN_IN(x1)
U51(x1, x2, x3)  =  U51(x3)
LESS_IN(x1, x2)  =  LESS_IN
U41(x1, x2, x3, x4, x5)  =  U41(x5)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

We have to consider all (P,R,Pi)-chains

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

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

IN_IN(X, tree(Y, X1, Right)) → U31(X, Y, X1, Right, less_in(Y, X))
IN_IN(X, tree(Y, X1, Right)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U51(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U31(X, Y, X1, Right, less_out(Y, X)) → U41(X, Y, X1, Right, in_in(X, Right))
U31(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
IN_IN(X, tree(Y, Left, X1)) → U11(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, Left, X1)) → LESS_IN(X, Y)
U11(X, Y, Left, X1, less_out(X, Y)) → U21(X, Y, Left, X1, in_in(X, Left))
U11(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
in_out(x1, x2)  =  in_out
IN_IN(x1, x2)  =  IN_IN(x1)
U51(x1, x2, x3)  =  U51(x3)
LESS_IN(x1, x2)  =  LESS_IN
U41(x1, x2, x3, x4, x5)  =  U41(x5)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

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

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
in_out(x1, x2)  =  in_out
LESS_IN(x1, x2)  =  LESS_IN

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
  ↳ PrologToPiTRSProof

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

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

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
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

LESS_IN → LESS_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

LESS_IN → LESS_IN

The TRS R consists of the following rules:none


s = LESS_IN evaluates to t =LESS_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from LESS_IN to LESS_IN.





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

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

IN_IN(X, tree(Y, Left, X1)) → U11(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, X1, Right)) → U31(X, Y, X1, Right, less_in(Y, X))
U31(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
U11(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
in_out(x1, x2)  =  in_out
IN_IN(x1, x2)  =  IN_IN(x1)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

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
  ↳ PrologToPiTRSProof

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

IN_IN(X, tree(Y, Left, X1)) → U11(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, X1, Right)) → U31(X, Y, X1, Right, less_in(Y, X))
U31(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
U11(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
IN_IN(x1, x2)  =  IN_IN(x1)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

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
                        ↳ Narrowing
  ↳ PrologToPiTRSProof

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

U11(X, less_out(X)) → IN_IN(X)
IN_IN(X) → U11(X, less_in)
U31(X, less_out(Y)) → IN_IN(X)
IN_IN(X) → U31(X, less_in)

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U5(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule IN_IN(X) → U11(X, less_in) at position [1] we obtained the following new rules:

IN_IN(y0) → U11(y0, U5(less_in))
IN_IN(y0) → U11(y0, less_out(0))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
  ↳ PrologToPiTRSProof

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

U11(X, less_out(X)) → IN_IN(X)
IN_IN(y0) → U11(y0, less_out(0))
U31(X, less_out(Y)) → IN_IN(X)
IN_IN(y0) → U11(y0, U5(less_in))
IN_IN(X) → U31(X, less_in)

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U5(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule IN_IN(X) → U31(X, less_in) at position [1] we obtained the following new rules:

IN_IN(y0) → U31(y0, less_out(0))
IN_IN(y0) → U31(y0, U5(less_in))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

IN_IN(y0) → U31(y0, less_out(0))
U11(X, less_out(X)) → IN_IN(X)
IN_IN(y0) → U11(y0, less_out(0))
U31(X, less_out(Y)) → IN_IN(X)
IN_IN(y0) → U11(y0, U5(less_in))
IN_IN(y0) → U31(y0, U5(less_in))

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U5(x0)

We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the right:

The TRS P consists of the following rules:

IN_IN(y0) → U31(y0, less_out(0))
U11(X, less_out(X)) → IN_IN(X)
IN_IN(y0) → U11(y0, less_out(0))
U31(X, less_out(Y)) → IN_IN(X)
IN_IN(y0) → U11(y0, U5(less_in))
IN_IN(y0) → U31(y0, U5(less_in))

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))


s = IN_IN(X) evaluates to t =IN_IN(X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

IN_IN(X) → U31(X, less_out(0))
with rule IN_IN(y0) → U31(y0, less_out(0)) and matcher [y0 / X].

U31(X, less_out(0)) → IN_IN(X)
with rule U31(X', less_out(Y)) → IN_IN(X') at position [] and matcher [Y / 0, X' / X]

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
in_out(x1, x2)  =  in_out(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
in_out(x1, x2)  =  in_out(x1)


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

IN_IN(X, tree(Y, X1, Right)) → U31(X, Y, X1, Right, less_in(Y, X))
IN_IN(X, tree(Y, X1, Right)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U51(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U31(X, Y, X1, Right, less_out(Y, X)) → U41(X, Y, X1, Right, in_in(X, Right))
U31(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
IN_IN(X, tree(Y, Left, X1)) → U11(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, Left, X1)) → LESS_IN(X, Y)
U11(X, Y, Left, X1, less_out(X, Y)) → U21(X, Y, Left, X1, in_in(X, Left))
U11(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
in_out(x1, x2)  =  in_out(x1)
IN_IN(x1, x2)  =  IN_IN(x1)
U51(x1, x2, x3)  =  U51(x3)
LESS_IN(x1, x2)  =  LESS_IN
U41(x1, x2, x3, x4, x5)  =  U41(x1, x5)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
U21(x1, x2, x3, x4, x5)  =  U21(x1, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

We have to consider all (P,R,Pi)-chains

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

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

IN_IN(X, tree(Y, X1, Right)) → U31(X, Y, X1, Right, less_in(Y, X))
IN_IN(X, tree(Y, X1, Right)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U51(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U31(X, Y, X1, Right, less_out(Y, X)) → U41(X, Y, X1, Right, in_in(X, Right))
U31(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
IN_IN(X, tree(Y, Left, X1)) → U11(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, Left, X1)) → LESS_IN(X, Y)
U11(X, Y, Left, X1, less_out(X, Y)) → U21(X, Y, Left, X1, in_in(X, Left))
U11(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
in_out(x1, x2)  =  in_out(x1)
IN_IN(x1, x2)  =  IN_IN(x1)
U51(x1, x2, x3)  =  U51(x3)
LESS_IN(x1, x2)  =  LESS_IN
U41(x1, x2, x3, x4, x5)  =  U41(x1, x5)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
U21(x1, x2, x3, x4, x5)  =  U21(x1, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

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

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
in_out(x1, x2)  =  in_out(x1)
LESS_IN(x1, x2)  =  LESS_IN

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

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

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

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP

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

LESS_IN → LESS_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

LESS_IN → LESS_IN

The TRS R consists of the following rules:none


s = LESS_IN evaluates to t =LESS_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from LESS_IN to LESS_IN.





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

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

IN_IN(X, tree(Y, Left, X1)) → U11(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, X1, Right)) → U31(X, Y, X1, Right, less_in(Y, X))
U31(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
U11(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2)  =  in_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U4(x1, x2, x3, x4, x5)  =  U4(x1, x5)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
in_out(x1, x2)  =  in_out(x1)
IN_IN(x1, x2)  =  IN_IN(x1)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

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

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

IN_IN(X, tree(Y, Left, X1)) → U11(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, X1, Right)) → U31(X, Y, X1, Right, less_in(Y, X))
U31(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
U11(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U5(x1, x2, x3)  =  U5(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
IN_IN(x1, x2)  =  IN_IN(x1)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing

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

U11(X, less_out(X)) → IN_IN(X)
IN_IN(X) → U11(X, less_in)
U31(X, less_out(Y)) → IN_IN(X)
IN_IN(X) → U31(X, less_in)

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U5(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule IN_IN(X) → U11(X, less_in) at position [1] we obtained the following new rules:

IN_IN(y0) → U11(y0, U5(less_in))
IN_IN(y0) → U11(y0, less_out(0))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing

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

U11(X, less_out(X)) → IN_IN(X)
IN_IN(y0) → U11(y0, less_out(0))
U31(X, less_out(Y)) → IN_IN(X)
IN_IN(y0) → U11(y0, U5(less_in))
IN_IN(X) → U31(X, less_in)

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U5(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule IN_IN(X) → U31(X, less_in) at position [1] we obtained the following new rules:

IN_IN(y0) → U31(y0, less_out(0))
IN_IN(y0) → U31(y0, U5(less_in))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ NonTerminationProof

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

IN_IN(y0) → U31(y0, less_out(0))
U11(X, less_out(X)) → IN_IN(X)
IN_IN(y0) → U11(y0, less_out(0))
U31(X, less_out(Y)) → IN_IN(X)
IN_IN(y0) → U11(y0, U5(less_in))
IN_IN(y0) → U31(y0, U5(less_in))

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U5(x0)

We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the right:

The TRS P consists of the following rules:

IN_IN(y0) → U31(y0, less_out(0))
U11(X, less_out(X)) → IN_IN(X)
IN_IN(y0) → U11(y0, less_out(0))
U31(X, less_out(Y)) → IN_IN(X)
IN_IN(y0) → U11(y0, U5(less_in))
IN_IN(y0) → U31(y0, U5(less_in))

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))


s = IN_IN(X) evaluates to t =IN_IN(X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

IN_IN(X) → U31(X, less_out(0))
with rule IN_IN(y0) → U31(y0, less_out(0)) and matcher [y0 / X].

U31(X, less_out(0)) → IN_IN(X)
with rule U31(X', less_out(Y)) → IN_IN(X') at position [] and matcher [Y / 0, X' / X]

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.