Certification Problem

Input (TPDB TRS_Standard/Rubio_04/lescanne)

The rewrite relation of the following TRS is considered.

div(X,e) i(X) (1)
i(div(X,Y)) div(Y,X) (2)
div(div(X,Y),Z) div(Y,div(i(X),Z)) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(i) = 3 weight(i) = 0
prec(div) = 0 weight(div) = 0
prec(e) = 2 weight(e) = 1
all of the following rules can be deleted.
div(X,e) i(X) (1)
i(div(X,Y)) div(Y,X) (2)
div(div(X,Y),Z) div(Y,div(i(X),Z)) (3)

1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.