Certification Problem

Input (TPDB TRS_Standard/SK90/2.17)

The rewrite relation of the following TRS is considered.

sum(0) → 0 (1)
sum(s(x)) → +(sum(x),s(x)) (2)
sum1(0) → 0 (3)
sum1(s(x)) → s(+(sum1(x),+(x,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 (3 x 3)-matrices with strict dimension 1 over the naturals
[sum1(x1)] =
1 1 1
1 1 1
1 1 1
· x1 +
0 0 0
1 0 0
1 0 0
[sum(x1)] =
1 1 0
0 0 0
1 1 0
· x1 +
0 0 0
0 0 0
1 0 0
[+(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
1 0 0
[0] =
0 0 0
0 0 0
1 0 0
[s(x1)] =
1 0 0
1 1 1
1 0 1
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
sum1(0) → 0 (3)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[sum1(x1)] =
1 1 1
1 1 1
1 1 1
· x1 +
0 0 0
1 0 0
1 0 0
[sum(x1)] =
1 1 1
1 0 1
1 1 0
· x1 +
1 0 0
1 0 0
0 0 0
[+(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 1
0 0 0
· x2 +
0 0 0
1 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
1 0 1
1 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
sum(0) → 0 (1)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[sum1(x1)] =
1 1 1
1 1 1
1 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[sum(x1)] =
1 1 0
1 1 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[+(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 1
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 1 0
1 1 1
1 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
sum1(s(x)) → s(+(sum1(x),+(x,x))) (4)

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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