Left Termination of the query pattern delmin_in_3(a, a, g) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇐)
    ↳10 QDP
    ↳11 QDPSizeChangeProof (⇔)
    ↳12 TRUE
  ↳13 PrologToPiTRSProof (⇐)
    ↳14 PiTRS
    ↳15 DependencyPairsProof (⇔)
    ↳16 PiDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 PiDP
    ↳19 UsableRulesProof (⇔)
    ↳20 PiDP
    ↳21 PiDPToQDPProof (⇐)
    ↳22 QDP

(0) Obligation:

Clauses:

delete(X, tree(X, void, Right), Right).
delete(X, tree(X, Left, void), Left).
delete(X, tree(X, Left, Right), tree(Y, Left, Right1)) :- delmin(Right, Y, Right1).
delete(X, tree(Y, Left, Right), tree(Y, Left1, Right)) :- ','(less(X, Y), delete(X, Left, Left1)).
delete(X, tree(Y, Left, Right), tree(Y, Left, Right1)) :- ','(less(Y, X), delete(X, Right, Right1)).
delmin(tree(Y, void, Right), Y, Right).
delmin(tree(X, Left, X1), Y, tree(X, Left1, X2)) :- delmin(Left, Y, Left1).
less(0, s(X3)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

delmin(a,a,g).

(1) PrologToPiTRSProof (SOUND transformation)

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

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_AAG(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN_AAG(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)
DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)
U6_AAG(x1, x2, x3, x4, x5, x6, x7)  =  U6_AAG(x7)

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

(4) Obligation:

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

DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_AAG(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN_AAG(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)
DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)
U6_AAG(x1, x2, x3, x4, x5, x6, x7)  =  U6_AAG(x7)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(6) Obligation:

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

DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN_AAG(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)
DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)

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

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN_AAG(Left, Y, Left1)

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)

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

(9) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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

DELMIN_IN_AAG(tree(X, Left1, X2)) → DELMIN_IN_AAG(Left1)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(11) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • DELMIN_IN_AAG(tree(X, Left1, X2)) → DELMIN_IN_AAG(Left1)
    The graph contains the following edges 1 > 1

(12) TRUE

(13) PrologToPiTRSProof (SOUND transformation)

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

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag(x3)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x1, x5, x6, x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(14) Obligation:

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

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag(x3)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x1, x5, x6, x7)

(15) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_AAG(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN_AAG(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag(x3)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x1, x5, x6, x7)
DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)
U6_AAG(x1, x2, x3, x4, x5, x6, x7)  =  U6_AAG(x1, x5, x6, x7)

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

(16) Obligation:

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

DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_AAG(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN_AAG(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag(x3)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x1, x5, x6, x7)
DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)
U6_AAG(x1, x2, x3, x4, x5, x6, x7)  =  U6_AAG(x1, x5, x6, x7)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(18) Obligation:

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

DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN_AAG(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in_aag(tree(Y, void, Right), Y, Right) → delmin_out_aag(tree(Y, void, Right), Y, Right)
delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) → delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
delmin_out_aag(x1, x2, x3)  =  delmin_out_aag(x3)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x1, x5, x6, x7)
DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)

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

(19) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(20) Obligation:

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

DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN_AAG(Left, Y, Left1)

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)

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

(21) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(22) Obligation:

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

DELMIN_IN_AAG(tree(X, Left1, X2)) → DELMIN_IN_AAG(Left1)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.