Certification Problem

Input (TPDB TRS_Equational/AProVE_AC_04/AC23)

The rewrite relation of the following equational TRS is considered.

zero(S) S (1)
plus(x,S) x (2)
plus(zero(x),zero(y)) zero(plus(x,y)) (3)
plus(zero(x),un(y)) un(plus(x,y)) (4)
plus(un(x),un(y)) zero(plus(x,plus(y,un(S)))) (5)
times(x,S) S (6)
times(x,times(S,z)) times(S,z) (7)
times(x,zero(y)) zero(times(x,y)) (8)
times(x,times(zero(y),z)) times(zero(times(x,y)),z) (9)
times(x,un(y)) plus(x,zero(times(x,y))) (10)
times(x,times(un(y),z)) times(plus(x,zero(times(x,y))),z) (11)

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 · x2 + 1 · x1
[times(x1, x2)] = 1 · x2 + 1 · x1 + 1 · x1 · x2
[zero(x1)] = 1 · x1
[S] = 0
[un(x1)] = 2 + 1 · x1
the rules
plus(un(x),un(y)) zero(plus(x,plus(y,un(S)))) (5)
times(x,un(y)) plus(x,zero(times(x,y))) (10)
times(x,times(un(y),z)) times(plus(x,zero(times(x,y))),z) (11)
can be deleted.

1.1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[plus(x1, x2)] = 3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
[times(x1, x2)] = 2 + 2 · x2 + 2 · x1 + 1 · x1 · x2
[zero(x1)] = 2 + 1 · x1
[S] = 3
[un(x1)] = 2 + 2 · x1
the rules
zero(S) S (1)
plus(x,S) x (2)
plus(zero(x),zero(y)) zero(plus(x,y)) (3)
plus(zero(x),un(y)) un(plus(x,y)) (4)
times(x,S) S (6)
times(x,times(S,z)) times(S,z) (7)
times(x,zero(y)) zero(times(x,y)) (8)
times(x,times(zero(y),z)) times(zero(times(x,y)),z) (9)
can be deleted.

1.1.1 R is empty

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