Certification Problem

Input (TPDB TRS_Equational/AProVE_AC_04/AC16)

The rewrite relation of the following equational TRS is considered.

plus(x,0) → x (1)
plus(x,x) → x (2)
plus(T(x),x) → T(x) (3)
plus(T(plus(x,y)),x) → T(plus(x,y)) (4)
L(T(x)) → L(x) (5)
L(plus(T(y),x)) → plus(L(plus(x,y)),L(y)) (6)
T(T(x)) → T(x) (7)
T(plus(T(y),x)) → plus(T(plus(x,y)),T(y)) (8)

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)] = 3 + 1 · x2 + 1 · x1
[0] = 3
[T(x1)] = 2 + 3 · x1 + 2 · x1 · x1
[L(x1)] = 2 · x1 · x1
the rules
plus(x,0) → x (1)
plus(x,x) → x (2)
plus(T(x),x) → T(x) (3)
plus(T(plus(x,y)),x) → T(plus(x,y)) (4)
L(T(x)) → L(x) (5)
L(plus(T(y),x)) → plus(L(plus(x,y)),L(y)) (6)
T(T(x)) → T(x) (7)
T(plus(T(y),x)) → plus(T(plus(x,y)),T(y)) (8)
can be deleted.

1.1 R is empty

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