Certification Problem

Input (TPDB TRS_Equational/Mixed_AC/YWHM14_5)

The rewrite relation of the following equational TRS is considered.

plus(f(x),g(x)) → plus(g(x),plus(g(x),g(x))) (1)
f(x) → plus(g(x),a) (2)

Associative symbols: plus

Commutative symbols: plus

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[plus(x1, x2)] = 1 · x2 + 1 · x1
[f(x1)] = 2 + 2 · x1 + 2 · x1 · x1
[g(x1)] = 1 · x1 · x1
[a] = 0
the rules
plus(f(x),g(x)) → plus(g(x),plus(g(x),g(x))) (1)
f(x) → plus(g(x),a) (2)
can be deleted.

1.1 R is empty

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