Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.34)

The rewrite relation of the following TRS is considered.

f(0) → true (1)
f(1) → false (2)
f(s(x)) → f(x) (3)
if(true,x,y) → x (4)
if(false,x,y) → y (5)
g(s(x),s(y)) → if(f(x),s(x),s(y)) (6)
g(x,c(y)) → g(x,g(s(c(y)),y)) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals
[false] =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[c(x1)] =
1 0 0 0
1 0 0 0
1 1 1 0
0 0 0 0
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[f(x1)] =
1 0 0 0
0 0 0 0
1 0 0 0
0 0 0 0
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
[1] =
1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[s(x1)] =
1 0 0 0
0 0 0 0
0 0 0 0
1 0 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
[g(x1, x2)] =
1 0 0 1
0 0 0 0
0 0 0 0
1 0 0 1
· x1 +
1 1 1 0
0 0 0 0
0 0 0 0
1 1 1 0
· x2 +
0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
[if(x1, x2, x3)] =
1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
· x1 +
1 0 0 0
0 1 0 0
0 0 1 0
1 0 0 1
· x2 +
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
· x3 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[true] =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
all of the following rules can be deleted.
f(1) → false (2)

1.1 Rule Removal

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals
[false] =
0 0 0 0
1 0 0 0
0 0 0 0
0 0 0 0
[c(x1)] =
1 0 0 0
1 1 1 1
1 1 0 0
0 1 0 0
· x1 +
0 0 0 0
1 0 0 0
0 0 0 0
0 0 0 0
[f(x1)] =
1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 1
· x1 +
0 0 0 0
0 0 0 0
1 0 0 0
0 0 0 0
[s(x1)] =
1 0 0 0
0 0 0 0
0 0 0 0
1 0 0 1
· 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
1 0 0 0
[g(x1, x2)] =
1 0 0 1
0 0 0 0
0 0 0 0
0 0 0 1
· x1 +
1 1 1 0
0 0 0 0
0 0 0 0
0 1 0 1
· x2 +
0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
[if(x1, x2, x3)] =
1 0 0 1
0 0 0 0
0 1 0 0
0 0 0 0
· x1 +
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
· x2 +
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
· x3 +
0 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
[true] =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
all of the following rules can be deleted.
g(x,c(y)) → g(x,g(s(c(y)),y)) (7)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[false] =
0 0 0
1 0 0
1 0 0
[f(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 1
1 0 1
0 1 1
· x1 +
1 0 0
0 0 0
1 0 0
[0] =
1 0 0
0 0 0
1 0 0
[g(x1, x2)] =
1 1 1
1 0 1
0 0 1
· x1 +
1 1 0
0 1 0
0 0 1
· x2 +
0 0 0
1 0 0
0 0 0
[if(x1, x2, x3)] =
1 0 0
0 1 1
0 0 0
· x1 +
1 0 1
0 1 0
0 0 1
· x2 +
1 1 0
0 1 0
0 0 1
· x3 +
0 0 0
0 0 0
0 0 0
[true] =
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(0) → true (1)
f(s(x)) → f(x) (3)
if(true,x,y) → x (4)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[false] =
0 0 0
1 0 0
1 0 0
[f(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[s(x1)] =
1 1 0
0 0 0
1 1 0
· x1 +
1 0 0
0 0 0
0 0 0
[g(x1, x2)] =
1 0 1
0 0 0
1 0 0
· x1 +
1 0 1
1 0 0
1 0 1
· x2 +
1 0 0
0 0 0
1 0 0
[if(x1, x2, x3)] =
1 1 0
0 0 1
0 1 1
· x1 +
1 0 0
0 0 0
0 0 1
· x2 +
1 0 1
0 1 1
1 0 1
· x3 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
if(false,x,y) → y (5)

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 1 0
0 0 0
1 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[s(x1)] =
1 0 1
1 1 0
1 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[g(x1, x2)] =
1 1 1
1 0 1
1 0 1
· x1 +
1 1 1
1 0 0
1 0 0
· x2 +
1 0 0
0 0 0
0 0 0
[if(x1, x2, x3)] =
1 0 0
0 0 1
0 0 1
· x1 +
1 0 1
1 0 0
1 0 0
· x2 +
1 1 1
1 0 0
1 0 0
· x3 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(s(x),s(y)) → if(f(x),s(x),s(y)) (6)

1.1.1.1.1.1 R is empty

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