Certification Problem

Input (TPDB TRS_Standard/SK90/2.34)

The rewrite relation of the following TRS is considered.

if(true,x,y) → x (1)
if(false,x,y) → y (2)
if(x,y,y) → y (3)
if(if(x,y,z),u,v) → if(x,if(y,u,v),if(z,u,v)) (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
[v] =
0 0 0
0 0 0
0 0 0
[if(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 1 0
0 0 1
· x2 +
1 0 0
0 1 0
0 0 1
· x3 +
0 0 0
0 0 0
0 0 0
[u] =
0 0 0
0 0 0
0 0 0
[true] =
0 0 0
0 0 0
0 0 0
[false] =
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
if(false,x,y) → y (2)

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(if) = 0 status(if) = [1, 3, 2] list-extension(if) = Lex
and the following Max-polynomial interpretation
[if(x1, x2, x3)] = max(0, 0 + 1 · x1, 0 + 1 · x2, 2 + 1 · x3)
all of the following rules can be deleted.
if(x,y,y) → y (3)

1.1.1.1.1 R is empty

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