Certification Problem

Input (TPDB TRS_Standard/SK90/4.12)

The rewrite relation of the following TRS is considered.

+(0,y) → y (1)
+(s(x),0) → s(x) (2)
+(s(x),s(y)) → s(+(s(x),+(y,0))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[+(x1, x2)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 1 0
1 1 0
0 0 1
· x2 +
0 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
+(s(x),s(y)) → s(+(s(x),+(y,0))) (3)

1.1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(s) = 0 status(s) = [1] list-extension(s) = Lex
prec(+) = 0 status(+) = [1, 2] list-extension(+) = Lex
prec(0) = 0 status(0) = [] list-extension(0) = Lex
and the following Max-polynomial interpretation
[s(x1)] = max(4, 4 + 1 · x1)
[+(x1, x2)] = max(0, 0 + 1 · x1, 4 + 1 · x2)
[0] = max(4)
all of the following rules can be deleted.
+(0,y) → y (1)
+(s(x),0) → s(x) (2)

1.1.1 R is empty

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