Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.28)

The rewrite relation of the following TRS is considered.

half(0) → 0 (1)
half(s(0)) → 0 (2)
half(s(s(x))) → s(half(x)) (3)
bits(0) → 0 (4)
bits(s(x)) → s(bits(half(s(x)))) (5)

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
[half(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[bits(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
bits(0) → 0 (4)

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

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

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[half(x1)] =
1 0 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[bits(x1)] =
1 1 0
0 0 1
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
[s(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
bits(s(x)) → s(bits(half(s(x)))) (5)

1.1.1.1.1.1 R is empty

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