Certification Problem

Input (TPDB TRS_Standard/SK90/4.43)

The rewrite relation of the following TRS is considered.

+(x,0) → x (1)
+(x,s(y)) → s(+(x,y)) (2)
+(0,y) → y (3)
+(s(x),y) → s(+(x,y)) (4)
+(x,+(y,z)) → +(+(x,y),z) (5)
f(g(f(x))) → f(h(s(0),x)) (6)
f(g(h(x,y))) → f(h(s(x),y)) (7)
f(h(x,h(y,z))) → f(h(+(x,y),z)) (8)

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 0 1
1 1 1
1 1 0
· x1 +
1 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
1 0 1
· x2 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 1
1 0 0
0 0 1
· x1 +
0 0 0
1 0 0
1 0 0
[h(x1, x2)] =
1 0 0
1 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[0] =
1 0 0
1 0 0
0 0 0
[s(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
+(x,0) → x (1)
+(0,y) → y (3)
f(g(f(x))) → f(h(s(0),x)) (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 0 1
1 1 0
· x1 +
1 0 0
1 0 0
1 0 0
[+(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 1 1
0 0 0
· x2 +
0 0 0
1 0 0
0 0 0
[f(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
1 0 0
1 0 0
[h(x1, x2)] =
1 0 1
0 0 0
1 0 0
· x1 +
1 0 1
0 0 0
1 0 1
· x2 +
0 0 0
1 0 0
1 0 0
[s(x1)] =
1 0 0
0 0 1
0 0 1
· x1 +
1 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
f(g(h(x,y))) → f(h(s(x),y)) (7)
f(h(x,h(y,z))) → f(h(+(x,y),z)) (8)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[+(x1, x2)] =
1 0 1
0 1 0
0 0 0
· x1 +
1 1 0
0 1 0
0 0 1
· x2 +
1 0 0
1 0 0
0 0 0
[s(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
+(x,+(y,z)) → +(+(x,y),z) (5)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(s) = 0 weight(s) = 2
prec(+) = 1 weight(+) = 0
all of the following rules can be deleted.
+(x,s(y)) → s(+(x,y)) (2)
+(s(x),y) → s(+(x,y)) (4)

1.1.1.1.1 R is empty

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