Certification Problem

Input (TPDB TRS_Standard/AG01/#3.37)

The rewrite relation of the following TRS is considered.

not(true) → false (1)
not(false) → true (2)
evenodd(x,0) → not(evenodd(x,s(0))) (3)
evenodd(0,s(0)) → false (4)
evenodd(s(x),s(0)) → evenodd(x,0) (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
[evenodd(x1, x2)] =
1 1 0
1 0 1
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
1 0 0
0 0 0
[not(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
0 0 0
1 0 0
0 0 0
[s(x1)] =
1 0 1
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[true] =
0 0 0
0 0 0
0 0 0
[false] =
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
evenodd(0,s(0)) → false (4)

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[evenodd(x1, x2)] =
1 0 0 1 0
0 0 0 0 0
1 0 0 0 0
1 0 1 0 0
1 0 0 0 0
· x1 +
1 0 0 0 1
0 0 0 0 0
0 0 1 0 1
0 0 0 0 1
0 0 0 0 0
· x2 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[not(x1)] =
1 1 0 0 0
0 1 0 0 0
0 0 0 0 1
0 0 1 0 0
0 0 1 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[0] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[s(x1)] =
1 1 1 1 0
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
[true] =
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
[false] =
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
not(true) → false (1)
not(false) → true (2)
evenodd(x,0) → not(evenodd(x,s(0))) (3)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(s) = 3 weight(s) = 2
prec(evenodd) = 0 weight(evenodd) = 0
prec(0) = 2 weight(0) = 1
all of the following rules can be deleted.
evenodd(s(x),s(0)) → evenodd(x,0) (5)

1.1.1.1 R is empty

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