Certification Problem

Input (TPDB TRS_Standard/Various_04/23)

The rewrite relation of the following TRS is considered.

g(0,f(x,x)) → x (1)
g(x,s(y)) → g(f(x,y),0) (2)
g(s(x),y) → g(f(x,y),0) (3)
g(f(x,y),0) → f(g(x,0),g(y,0)) (4)

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
[f(x1, x2)] =
1 0 0
0 0 1
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 1 0
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
1 0 0
[0] =
0 0 0
0 0 0
0 0 0
[g(x1, x2)] =
1 0 0
1 0 0
1 0 0
· x1 +
1 0 1
1 0 1
1 1 0
· x2 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(x,s(y)) → g(f(x,y),0) (2)

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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