Certification Problem

Input (TPDB TRS_Standard/Rubio_04/bn129)

The rewrite relation of the following TRS is considered.

plus(s(X),plus(Y,Z)) → plus(X,plus(s(s(Y)),Z)) (1)
plus(s(X1),plus(X2,plus(X3,X4))) → plus(X1,plus(X3,plus(X2,X4))) (2)

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(plus) = 0 weight(plus) = 0
prec(s) = 1 weight(s) = 0
all of the following rules can be deleted.
plus(s(X),plus(Y,Z)) → plus(X,plus(s(s(Y)),Z)) (1)
plus(s(X1),plus(X2,plus(X3,X4))) → plus(X1,plus(X3,plus(X2,X4))) (2)

1.1 R is empty

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