Certification Problem

Input (TPDB TRS_Standard/HirokawaMiddeldorp_04/t013)

The rewrite relation of the following TRS is considered.

-(x,0) → x (1)
-(0,s(y)) → 0 (2)
-(s(x),s(y)) → -(x,y) (3)
f(0) → 0 (4)
f(s(x)) → -(s(x),g(f(x))) (5)
g(0) → s(0) (6)
g(s(x)) → -(s(x),f(g(x))) (7)

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
[g(x1)] =
1 1 0
0 0 1
0 1 1
· x1 +
1 0 0
1 0 0
1 0 0
[-(x1, x2)] =
1 0 0
0 1 0
0 1 1
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 1 0
0 0 1
0 1 1
· x1 +
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 1 1
1 1 1
· x1 +
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
g(0) → s(0) (6)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[g(x1)] = 4 · x1 + 2
[-(x1, x2)] = 2 · x1 + 2 · x2 + 0
[f(x1)] = 4 · x1 + 0
[0] = 4
[s(x1)] = 16 · x1 + 7
all of the following rules can be deleted.
-(x,0) → x (1)
-(0,s(y)) → 0 (2)
-(s(x),s(y)) → -(x,y) (3)
f(s(x)) → -(s(x),g(f(x))) (5)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[g(x1)] = 16 · x1 + 0
[-(x1, x2)] = 8 · x1 + 1 · x2 + 7
[f(x1)] = 4 · x1 + 17
[s(x1)] = 8 · x1 + 4
all of the following rules can be deleted.
g(s(x)) → -(s(x),f(g(x))) (7)

1.1.1.1.1 R is empty

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