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

0 Prolog
↳1 PrologToPrologProblemTransformerProof (⇐)
↳2 Prolog
↳3 PrologToPiTRSProof (⇐)
↳4 PiTRS
↳5 DependencyPairsProof (⇔)
↳6 PiDP
↳7 DependencyGraphProof (⇔)
↳8 PiDP
↳9 UsableRulesProof (⇔)
↳10 PiDP
↳11 PiDPToQDPProof (⇐)
↳12 QDP
↳13 QDPSizeChangeProof (⇔)
↳14 YES

(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(g,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

delmin1(tree(T6, void, T7), T6, T7).
delmin1(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31)).
delmin1(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) :- delmin1(T57, T62, T63).

Queries:

delmin1(g,a,a).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
delmin1_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

delmin1_in_gaa(tree(T6, void, T7), T6, T7) → delmin1_out_gaa(tree(T6, void, T7), T6, T7)
delmin1_in_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31)) → delmin1_out_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31))
delmin1_in_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_in_gaa(T57, T62, T63))
U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_out_gaa(T57, T62, T63)) → delmin1_out_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31))

The argument filtering Pi contains the following mapping:
delmin1_in_gaa(x1, x2, x3)  =  delmin1_in_gaa(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
void  =  void
delmin1_out_gaa(x1, x2, x3)  =  delmin1_out_gaa(x2)
U1_gaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)  =  U1_gaa(x10)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

delmin1_in_gaa(tree(T6, void, T7), T6, T7) → delmin1_out_gaa(tree(T6, void, T7), T6, T7)
delmin1_in_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31)) → delmin1_out_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31))
delmin1_in_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_in_gaa(T57, T62, T63))
U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_out_gaa(T57, T62, T63)) → delmin1_out_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31))

The argument filtering Pi contains the following mapping:
delmin1_in_gaa(x1, x2, x3)  =  delmin1_in_gaa(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
void  =  void
delmin1_out_gaa(x1, x2, x3)  =  delmin1_out_gaa(x2)
U1_gaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)  =  U1_gaa(x10)

(5) 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:

DELMIN1_IN_GAA(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → U1_GAA(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_in_gaa(T57, T62, T63))
DELMIN1_IN_GAA(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → DELMIN1_IN_GAA(T57, T62, T63)

The TRS R consists of the following rules:

delmin1_in_gaa(tree(T6, void, T7), T6, T7) → delmin1_out_gaa(tree(T6, void, T7), T6, T7)
delmin1_in_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31)) → delmin1_out_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31))
delmin1_in_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_in_gaa(T57, T62, T63))
U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_out_gaa(T57, T62, T63)) → delmin1_out_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31))

The argument filtering Pi contains the following mapping:
delmin1_in_gaa(x1, x2, x3)  =  delmin1_in_gaa(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
void  =  void
delmin1_out_gaa(x1, x2, x3)  =  delmin1_out_gaa(x2)
U1_gaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)  =  U1_gaa(x10)
DELMIN1_IN_GAA(x1, x2, x3)  =  DELMIN1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)  =  U1_GAA(x10)

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

(6) Obligation:

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

DELMIN1_IN_GAA(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → U1_GAA(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_in_gaa(T57, T62, T63))
DELMIN1_IN_GAA(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → DELMIN1_IN_GAA(T57, T62, T63)

The TRS R consists of the following rules:

delmin1_in_gaa(tree(T6, void, T7), T6, T7) → delmin1_out_gaa(tree(T6, void, T7), T6, T7)
delmin1_in_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31)) → delmin1_out_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31))
delmin1_in_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_in_gaa(T57, T62, T63))
U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_out_gaa(T57, T62, T63)) → delmin1_out_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31))

The argument filtering Pi contains the following mapping:
delmin1_in_gaa(x1, x2, x3)  =  delmin1_in_gaa(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
void  =  void
delmin1_out_gaa(x1, x2, x3)  =  delmin1_out_gaa(x2)
U1_gaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)  =  U1_gaa(x10)
DELMIN1_IN_GAA(x1, x2, x3)  =  DELMIN1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)  =  U1_GAA(x10)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

DELMIN1_IN_GAA(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → DELMIN1_IN_GAA(T57, T62, T63)

The TRS R consists of the following rules:

delmin1_in_gaa(tree(T6, void, T7), T6, T7) → delmin1_out_gaa(tree(T6, void, T7), T6, T7)
delmin1_in_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31)) → delmin1_out_gaa(tree(T26, tree(T42, void, T43), T28), T42, tree(T26, T43, T31))
delmin1_in_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_in_gaa(T57, T62, T63))
U1_gaa(T26, T56, T57, T58, T28, T62, T63, T61, T31, delmin1_out_gaa(T57, T62, T63)) → delmin1_out_gaa(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31))

The argument filtering Pi contains the following mapping:
delmin1_in_gaa(x1, x2, x3)  =  delmin1_in_gaa(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
void  =  void
delmin1_out_gaa(x1, x2, x3)  =  delmin1_out_gaa(x2)
U1_gaa(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)  =  U1_gaa(x10)
DELMIN1_IN_GAA(x1, x2, x3)  =  DELMIN1_IN_GAA(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

DELMIN1_IN_GAA(tree(T26, tree(T56, T57, T58), T28), T62, tree(T26, tree(T56, T63, T61), T31)) → DELMIN1_IN_GAA(T57, T62, T63)

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
DELMIN1_IN_GAA(x1, x2, x3)  =  DELMIN1_IN_GAA(x1)

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

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

DELMIN1_IN_GAA(tree(T26, tree(T56, T57, T58), T28)) → DELMIN1_IN_GAA(T57)

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

(13) 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:

  • DELMIN1_IN_GAA(tree(T26, tree(T56, T57, T58), T28)) → DELMIN1_IN_GAA(T57)
    The graph contains the following edges 1 > 1

(14) YES