Certification Problem

Input (TPDB TRS_Equational/AProVE_AC_04/AC14)

The rewrite relation of the following equational TRS is considered.

plus(x,0) x (1)
plus(x,s(y)) s(plus(x,y)) (2)
times(x,0) 0 (3)
times(x,s(y)) plus(times(x,y),x) (4)
times(x,plus(y,z)) plus(times(x,y),times(x,z)) (5)

Associative symbols: plus, times

Commutative symbols: plus, times

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 + 1 · x2 + 1 · x1
[times(x1, x2)] = 1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
[0] = 0
[s(x1)] = 2 + 1 · x1
the rules
plus(x,0) x (1)
times(x,0) 0 (3)
times(x,s(y)) plus(times(x,y),x) (4)
can be deleted.

1.1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[plus(x1, x2)] = 2 + 1 · x2 + 1 · x1
[times(x1, x2)] = 2 + 3 · x2 + 3 · x1 + 3 · x1 · x2
[s(x1)] = 1 · x1
the rule
times(x,plus(y,z)) plus(times(x,y),times(x,z)) (5)
can be deleted.

1.1.1 AC Rule Removal

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

1.1.1.1 R is empty

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