Certification Problem

Input (TPDB TRS_Standard/SK90/2.20)

The rewrite relation of the following TRS is considered.

sum(0) → 0 (1)
sum(s(x)) → +(sqr(s(x)),sum(x)) (2)
sqr(x) → *(x,x) (3)
sum(s(x)) → +(*(s(x),s(x)),sum(x)) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[+(x1, x2)] =
5 0
4 0
· x1 +
1 0
0 0
· x2 +
0 0
0 0
[sum(x1)] =
2 2
4 1
· x1 +
0 0
0 0
[sqr(x1)] =
2 0
0 0
· x1 +
0 0
4 0
[*(x1, x2)] =
1 0
0 0
· x1 +
1 0
0 0
· x2 +
0 0
1 0
[0] =
2 0
4 0
[s(x1)] =
1 0
5 4
· x1 +
0 0
0 0
all of the following rules can be deleted.
sum(0) → 0 (1)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[+(x1, x2)] = 8 · x1 + 1 · x2 + 0
[sum(x1)] = 20 · x1 + 0
[sqr(x1)] = 2 · x1 + 4
[*(x1, x2)] = 1 · x1 + 1 · x2 + 2
[s(x1)] = 8 · x1 + 8
all of the following rules can be deleted.
sqr(x) → *(x,x) (3)
sum(s(x)) → +(*(s(x),s(x)),sum(x)) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[+(x1, x2)] =
1 0 1
1 0 1
1 0 0
· x1 +
1 0 0
0 0 0
1 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[sum(x1)] =
1 1 1
1 1 0
1 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[sqr(x1)] =
1 0 0
0 1 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 1 0
1 1 1
· x1 +
1 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
sum(s(x)) → +(sqr(s(x)),sum(x)) (2)

1.1.1.1 R is empty

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