Certification Problem

Input (TPDB TRS_Equational/AProVE_AC_04/AC12)

The rewrite relation of the following equational TRS is considered.

f(g(f(h(x),x)),x) f(h(x),f(x,x)) (1)
f(h(x),g(x)) f(g(h(x)),x) (2)
f(g(h(x)),f(x,f(x,y))) f(g(f(h(x),y)),x) (3)
f(g(g(x)),x) f(g(x),g(x)) (4)

Associative symbols: f

Commutative symbols: f

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
[f(x1, x2)] = 2 + 3 · x2 + 3 · x1 + 3 · x1 · x2
[g(x1)] = 1 + 3 · x1
[h(x1)] = 1 · x1
the rules
f(g(f(h(x),x)),x) f(h(x),f(x,x)) (1)
f(g(h(x)),f(x,f(x,y))) f(g(f(h(x),y)),x) (3)
f(g(g(x)),x) f(g(x),g(x)) (4)
can be deleted.

1.1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[f(x1, x2)] = 1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
[h(x1)] = 2 + 1 · x1
[g(x1)] = 3 + 2 · x1
the rule
f(h(x),g(x)) f(g(h(x)),x) (2)
can be deleted.

1.1.1 R is empty

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