Certification Problem

Input (TPDB TRS_Standard/SK90/2.01)

The rewrite relation of the following TRS is considered.

i(0) → 0 (1)
+(0,y) → y (2)
+(x,0) → x (3)
i(i(x)) → x (4)
+(i(x),x) → 0 (5)
+(x,i(x)) → 0 (6)
i(+(x,y)) → +(i(x),i(y)) (7)
+(x,+(y,z)) → +(+(x,y),z) (8)
+(+(x,i(y)),y) → x (9)
+(+(x,y),i(y)) → x (10)

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
[i(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
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 1 0
0 0 1
· x2 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
+(0,y) → y (2)
+(x,0) → x (3)
+(i(x),x) → 0 (5)
+(x,i(x)) → 0 (6)
+(+(x,i(y)),y) → x (9)
+(+(x,y),i(y)) → x (10)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[i(x1)] =
1 1 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
1 0 0
1 0 0
0 0 0
[+(x1, x2)] =
1 0 0
0 1 1
0 0 0
· x1 +
1 1 0
0 1 1
0 0 0
· x2 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
i(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
[i(x1)] =
1 0 0
1 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[+(x1, x2)] =
1 0 0
0 0 0
0 0 1
· x1 +
1 0 1
0 0 0
1 0 0
· x2 +
1 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
+(x,+(y,z)) → +(+(x,y),z) (8)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(+) = 0 weight(+) = 0
prec(i) = 1 weight(i) = 0
all of the following rules can be deleted.
i(i(x)) → x (4)
i(+(x,y)) → +(i(x),i(y)) (7)

1.1.1.1.1 R is empty

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