Left Termination of the query pattern minimum_in_2(a, g) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

minimum(tree(X, void, X1), X).
minimum(tree(X1, Left, X2), X) :- minimum(Left, X).

Queries:

minimum(a,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:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1)

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:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
void  =  void
minimum_out(x1, x2)  =  minimum_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:

MINIMUM_IN(tree(X1, Left, X2), X) → U11(X1, Left, X2, X, minimum_in(Left, X))
MINIMUM_IN(tree(X1, Left, X2), X) → MINIMUM_IN(Left, X)

The TRS R consists of the following rules:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1)
MINIMUM_IN(x1, x2)  =  MINIMUM_IN(x2)
U11(x1, x2, x3, x4, x5)  =  U11(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:

MINIMUM_IN(tree(X1, Left, X2), X) → U11(X1, Left, X2, X, minimum_in(Left, X))
MINIMUM_IN(tree(X1, Left, X2), X) → MINIMUM_IN(Left, X)

The TRS R consists of the following rules:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1)
MINIMUM_IN(x1, x2)  =  MINIMUM_IN(x2)
U11(x1, x2, x3, x4, x5)  =  U11(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

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

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

MINIMUM_IN(tree(X1, Left, X2), X) → MINIMUM_IN(Left, X)

The TRS R consists of the following rules:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1)
MINIMUM_IN(x1, x2)  =  MINIMUM_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

MINIMUM_IN(tree(X1, Left, X2), X) → MINIMUM_IN(Left, X)

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x2)
MINIMUM_IN(x1, x2)  =  MINIMUM_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

MINIMUM_IN(X) → MINIMUM_IN(X)

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:

MINIMUM_IN(X) → MINIMUM_IN(X)

The TRS R consists of the following rules:none


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

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 MINIMUM_IN(X) to MINIMUM_IN(X).




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:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1, x2)

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:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1, 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:

MINIMUM_IN(tree(X1, Left, X2), X) → U11(X1, Left, X2, X, minimum_in(Left, X))
MINIMUM_IN(tree(X1, Left, X2), X) → MINIMUM_IN(Left, X)

The TRS R consists of the following rules:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1, x2)
MINIMUM_IN(x1, x2)  =  MINIMUM_IN(x2)
U11(x1, x2, x3, x4, x5)  =  U11(x4, 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:

MINIMUM_IN(tree(X1, Left, X2), X) → U11(X1, Left, X2, X, minimum_in(Left, X))
MINIMUM_IN(tree(X1, Left, X2), X) → MINIMUM_IN(Left, X)

The TRS R consists of the following rules:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1, x2)
MINIMUM_IN(x1, x2)  =  MINIMUM_IN(x2)
U11(x1, x2, x3, x4, x5)  =  U11(x4, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

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

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

MINIMUM_IN(tree(X1, Left, X2), X) → MINIMUM_IN(Left, X)

The TRS R consists of the following rules:

minimum_in(tree(X1, Left, X2), X) → U1(X1, Left, X2, X, minimum_in(Left, X))
minimum_in(tree(X, void, X1), X) → minimum_out(tree(X, void, X1), X)
U1(X1, Left, X2, X, minimum_out(Left, X)) → minimum_out(tree(X1, Left, X2), X)

The argument filtering Pi contains the following mapping:
minimum_in(x1, x2)  =  minimum_in(x2)
tree(x1, x2, x3)  =  tree(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
void  =  void
minimum_out(x1, x2)  =  minimum_out(x1, x2)
MINIMUM_IN(x1, x2)  =  MINIMUM_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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

MINIMUM_IN(tree(X1, Left, X2), X) → MINIMUM_IN(Left, X)

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x2)
MINIMUM_IN(x1, x2)  =  MINIMUM_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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ NonTerminationProof

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

MINIMUM_IN(X) → MINIMUM_IN(X)

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:

MINIMUM_IN(X) → MINIMUM_IN(X)

The TRS R consists of the following rules:none


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

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 MINIMUM_IN(X) to MINIMUM_IN(X).