Left Termination of the query pattern delete_in_3(g, a, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

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(X)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

delete(g,a,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:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_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:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_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:

DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U41(X, Y, Left, Right, Right1, less_in(Y, X))
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U71(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → U51(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → DELETE_IN(X, Right, Right1)
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U21(X, Y, Left, Right, Left1, less_in(X, Y))
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → LESS_IN(X, Y)
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → U31(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → DELETE_IN(X, Left, Left1)
DELETE_IN(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U11(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
DELETE_IN(X, tree(X, Left, Right), tree(Y, Left, Right1)) → DELMIN_IN(Right, Y, Right1)
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U61(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN(Left, Y, Left1)

The TRS R consists of the following rules:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_out
DELETE_IN(x1, x2, x3)  =  DELETE_IN(x1)
U71(x1, x2, x3)  =  U71(x3)
LESS_IN(x1, x2)  =  LESS_IN
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
DELMIN_IN(x1, x2, x3)  =  DELMIN_IN
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x6)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x1, x6)
U11(x1, x2, x3, x4, x5, x6)  =  U11(x6)
U61(x1, x2, x3, x4, x5, x6, x7)  =  U61(x7)
U31(x1, x2, x3, x4, x5, x6)  =  U31(x6)

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:

DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U41(X, Y, Left, Right, Right1, less_in(Y, X))
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U71(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → U51(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → DELETE_IN(X, Right, Right1)
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U21(X, Y, Left, Right, Left1, less_in(X, Y))
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → LESS_IN(X, Y)
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → U31(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → DELETE_IN(X, Left, Left1)
DELETE_IN(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U11(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
DELETE_IN(X, tree(X, Left, Right), tree(Y, Left, Right1)) → DELMIN_IN(Right, Y, Right1)
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U61(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN(Left, Y, Left1)

The TRS R consists of the following rules:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_out
DELETE_IN(x1, x2, x3)  =  DELETE_IN(x1)
U71(x1, x2, x3)  =  U71(x3)
LESS_IN(x1, x2)  =  LESS_IN
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
DELMIN_IN(x1, x2, x3)  =  DELMIN_IN
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x6)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x1, x6)
U11(x1, x2, x3, x4, x5, x6)  =  U11(x6)
U61(x1, x2, x3, x4, x5, x6, x7)  =  U61(x7)
U31(x1, x2, x3, x4, x5, x6)  =  U31(x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 8 less nodes.

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

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

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

The TRS R consists of the following rules:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_out
DELMIN_IN(x1, x2, x3)  =  DELMIN_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
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

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

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
DELMIN_IN(x1, x2, x3)  =  DELMIN_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
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

DELMIN_IN → DELMIN_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:

DELMIN_IN → DELMIN_IN

The TRS R consists of the following rules:none


s = DELMIN_IN evaluates to t =DELMIN_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 DELMIN_IN to DELMIN_IN.





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

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_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
              ↳ 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
              ↳ 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
              ↳ PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U21(X, Y, Left, Right, Left1, less_in(X, Y))
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → DELETE_IN(X, Left, Left1)
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → DELETE_IN(X, Right, Right1)
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U41(X, Y, Left, Right, Right1, less_in(Y, X))

The TRS R consists of the following rules:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_out
DELETE_IN(x1, x2, x3)  =  DELETE_IN(x1)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x6)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x1, x6)

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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U21(X, Y, Left, Right, Left1, less_in(X, Y))
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → DELETE_IN(X, Left, Left1)
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → DELETE_IN(X, Right, Right1)
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U41(X, Y, Left, Right, Right1, less_in(Y, X))

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(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)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
DELETE_IN(x1, x2, x3)  =  DELETE_IN(x1)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x6)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x1, x6)

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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
  ↳ PrologToPiTRSProof

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

DELETE_IN(X) → U21(X, less_in)
DELETE_IN(X) → U41(X, less_in)
U21(X, less_out(X)) → DELETE_IN(X)
U41(X, less_out(Y)) → DELETE_IN(X)

The TRS R consists of the following rules:

less_in → U7(less_in)
less_in → less_out(0)
U7(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U7(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule DELETE_IN(X) → U21(X, less_in) at position [1] we obtained the following new rules:

DELETE_IN(y0) → U21(y0, U7(less_in))
DELETE_IN(y0) → U21(y0, less_out(0))



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

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

DELETE_IN(y0) → U21(y0, less_out(0))
DELETE_IN(y0) → U21(y0, U7(less_in))
DELETE_IN(X) → U41(X, less_in)
U21(X, less_out(X)) → DELETE_IN(X)
U41(X, less_out(Y)) → DELETE_IN(X)

The TRS R consists of the following rules:

less_in → U7(less_in)
less_in → less_out(0)
U7(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U7(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule DELETE_IN(X) → U41(X, less_in) at position [1] we obtained the following new rules:

DELETE_IN(y0) → U41(y0, less_out(0))
DELETE_IN(y0) → U41(y0, U7(less_in))



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

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

DELETE_IN(y0) → U41(y0, less_out(0))
DELETE_IN(y0) → U41(y0, U7(less_in))
DELETE_IN(y0) → U21(y0, less_out(0))
DELETE_IN(y0) → U21(y0, U7(less_in))
U21(X, less_out(X)) → DELETE_IN(X)
U41(X, less_out(Y)) → DELETE_IN(X)

The TRS R consists of the following rules:

less_in → U7(less_in)
less_in → less_out(0)
U7(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U7(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:

DELETE_IN(y0) → U41(y0, less_out(0))
DELETE_IN(y0) → U41(y0, U7(less_in))
DELETE_IN(y0) → U21(y0, less_out(0))
DELETE_IN(y0) → U21(y0, U7(less_in))
U21(X, less_out(X)) → DELETE_IN(X)
U41(X, less_out(Y)) → DELETE_IN(X)

The TRS R consists of the following rules:

less_in → U7(less_in)
less_in → less_out(0)
U7(less_out(X)) → less_out(s(X))


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

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



Rewriting sequence

DELETE_IN(X) → U41(X, less_out(0))
with rule DELETE_IN(y0) → U41(y0, less_out(0)) and matcher [y0 / X].

U41(X, less_out(0)) → DELETE_IN(X)
with rule U41(X', less_out(Y)) → DELETE_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:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x1, x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x1, x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x1, x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_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:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x1, x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x1, x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x1, x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_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:

DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U41(X, Y, Left, Right, Right1, less_in(Y, X))
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U71(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → U51(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → DELETE_IN(X, Right, Right1)
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U21(X, Y, Left, Right, Left1, less_in(X, Y))
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → LESS_IN(X, Y)
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → U31(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → DELETE_IN(X, Left, Left1)
DELETE_IN(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U11(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
DELETE_IN(X, tree(X, Left, Right), tree(Y, Left, Right1)) → DELMIN_IN(Right, Y, Right1)
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U61(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN(Left, Y, Left1)

The TRS R consists of the following rules:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x1, x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x1, x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x1, x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_out(x1)
DELETE_IN(x1, x2, x3)  =  DELETE_IN(x1)
U71(x1, x2, x3)  =  U71(x3)
LESS_IN(x1, x2)  =  LESS_IN
U51(x1, x2, x3, x4, x5, x6)  =  U51(x1, x6)
DELMIN_IN(x1, x2, x3)  =  DELMIN_IN
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x6)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x1, x6)
U11(x1, x2, x3, x4, x5, x6)  =  U11(x1, x6)
U61(x1, x2, x3, x4, x5, x6, x7)  =  U61(x7)
U31(x1, x2, x3, x4, x5, x6)  =  U31(x1, x6)

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:

DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U41(X, Y, Left, Right, Right1, less_in(Y, X))
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U71(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → U51(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → DELETE_IN(X, Right, Right1)
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U21(X, Y, Left, Right, Left1, less_in(X, Y))
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → LESS_IN(X, Y)
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → U31(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → DELETE_IN(X, Left, Left1)
DELETE_IN(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U11(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
DELETE_IN(X, tree(X, Left, Right), tree(Y, Left, Right1)) → DELMIN_IN(Right, Y, Right1)
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U61(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN(Left, Y, Left1)

The TRS R consists of the following rules:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x1, x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x1, x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x1, x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_out(x1)
DELETE_IN(x1, x2, x3)  =  DELETE_IN(x1)
U71(x1, x2, x3)  =  U71(x3)
LESS_IN(x1, x2)  =  LESS_IN
U51(x1, x2, x3, x4, x5, x6)  =  U51(x1, x6)
DELMIN_IN(x1, x2, x3)  =  DELMIN_IN
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x6)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x1, x6)
U11(x1, x2, x3, x4, x5, x6)  =  U11(x1, x6)
U61(x1, x2, x3, x4, x5, x6, x7)  =  U61(x7)
U31(x1, x2, x3, x4, x5, x6)  =  U31(x1, x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 8 less nodes.

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

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

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

The TRS R consists of the following rules:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x1, x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x1, x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x1, x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_out(x1)
DELMIN_IN(x1, x2, x3)  =  DELMIN_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
              ↳ PiDP

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

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

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

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

DELMIN_IN → DELMIN_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:

DELMIN_IN → DELMIN_IN

The TRS R consists of the following rules:none


s = DELMIN_IN evaluates to t =DELMIN_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 DELMIN_IN to DELMIN_IN.





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

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x1, x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x1, x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x1, x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_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
              ↳ 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
              ↳ 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
              ↳ PiDP
                ↳ UsableRulesProof

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

DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U21(X, Y, Left, Right, Left1, less_in(X, Y))
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → DELETE_IN(X, Left, Left1)
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → DELETE_IN(X, Right, Right1)
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U41(X, Y, Left, Right, Right1, less_in(Y, X))

The TRS R consists of the following rules:

delete_in(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U4(X, Y, Left, Right, Right1, less_in(Y, X))
less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U4(X, Y, Left, Right, Right1, less_out(Y, X)) → U5(X, Y, Left, Right, Right1, delete_in(X, Right, Right1))
delete_in(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U2(X, Y, Left, Right, Left1, less_in(X, Y))
U2(X, Y, Left, Right, Left1, less_out(X, Y)) → U3(X, Y, Left, Right, Left1, delete_in(X, Left, Left1))
delete_in(X, tree(X, Left, Right), tree(Y, Left, Right1)) → U1(X, Left, Right, Y, Right1, delmin_in(Right, Y, Right1))
delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))
U1(X, Left, Right, Y, Right1, delmin_out(Right, Y, Right1)) → delete_out(X, tree(X, Left, Right), tree(Y, Left, Right1))
delete_in(X, tree(X, Left, void), Left) → delete_out(X, tree(X, Left, void), Left)
delete_in(X, tree(X, void, Right), Right) → delete_out(X, tree(X, void, Right), Right)
U3(X, Y, Left, Right, Left1, delete_out(X, Left, Left1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left1, Right))
U5(X, Y, Left, Right, Right1, delete_out(X, Right, Right1)) → delete_out(X, tree(Y, Left, Right), tree(Y, Left, Right1))

The argument filtering Pi contains the following mapping:
delete_in(x1, x2, x3)  =  delete_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x1, x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x1, x6)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x1, x6)
U3(x1, x2, x3, x4, x5, x6)  =  U3(x1, x6)
U1(x1, x2, x3, x4, x5, x6)  =  U1(x1, x6)
delmin_in(x1, x2, x3)  =  delmin_in
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out
delete_out(x1, x2, x3)  =  delete_out(x1)
DELETE_IN(x1, x2, x3)  =  DELETE_IN(x1)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x6)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x1, x6)

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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left1, Right)) → U21(X, Y, Left, Right, Left1, less_in(X, Y))
U21(X, Y, Left, Right, Left1, less_out(X, Y)) → DELETE_IN(X, Left, Left1)
U41(X, Y, Left, Right, Right1, less_out(Y, X)) → DELETE_IN(X, Right, Right1)
DELETE_IN(X, tree(Y, Left, Right), tree(Y, Left, Right1)) → U41(X, Y, Left, Right, Right1, less_in(Y, X))

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U7(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U7(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)
U7(x1, x2, x3)  =  U7(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
DELETE_IN(x1, x2, x3)  =  DELETE_IN(x1)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x1, x6)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x1, x6)

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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing

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

DELETE_IN(X) → U21(X, less_in)
DELETE_IN(X) → U41(X, less_in)
U21(X, less_out(X)) → DELETE_IN(X)
U41(X, less_out(Y)) → DELETE_IN(X)

The TRS R consists of the following rules:

less_in → U7(less_in)
less_in → less_out(0)
U7(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U7(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule DELETE_IN(X) → U21(X, less_in) at position [1] we obtained the following new rules:

DELETE_IN(y0) → U21(y0, U7(less_in))
DELETE_IN(y0) → U21(y0, less_out(0))



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

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

DELETE_IN(y0) → U21(y0, less_out(0))
DELETE_IN(y0) → U21(y0, U7(less_in))
DELETE_IN(X) → U41(X, less_in)
U21(X, less_out(X)) → DELETE_IN(X)
U41(X, less_out(Y)) → DELETE_IN(X)

The TRS R consists of the following rules:

less_in → U7(less_in)
less_in → less_out(0)
U7(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U7(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule DELETE_IN(X) → U41(X, less_in) at position [1] we obtained the following new rules:

DELETE_IN(y0) → U41(y0, less_out(0))
DELETE_IN(y0) → U41(y0, U7(less_in))



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

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

DELETE_IN(y0) → U41(y0, less_out(0))
DELETE_IN(y0) → U41(y0, U7(less_in))
DELETE_IN(y0) → U21(y0, less_out(0))
DELETE_IN(y0) → U21(y0, U7(less_in))
U21(X, less_out(X)) → DELETE_IN(X)
U41(X, less_out(Y)) → DELETE_IN(X)

The TRS R consists of the following rules:

less_in → U7(less_in)
less_in → less_out(0)
U7(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U7(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:

DELETE_IN(y0) → U41(y0, less_out(0))
DELETE_IN(y0) → U41(y0, U7(less_in))
DELETE_IN(y0) → U21(y0, less_out(0))
DELETE_IN(y0) → U21(y0, U7(less_in))
U21(X, less_out(X)) → DELETE_IN(X)
U41(X, less_out(Y)) → DELETE_IN(X)

The TRS R consists of the following rules:

less_in → U7(less_in)
less_in → less_out(0)
U7(less_out(X)) → less_out(s(X))


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

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



Rewriting sequence

DELETE_IN(X) → U41(X, less_out(0))
with rule DELETE_IN(y0) → U41(y0, less_out(0)) and matcher [y0 / X].

U41(X, less_out(0)) → DELETE_IN(X)
with rule U41(X', less_out(Y)) → DELETE_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.