Certification Problem

Input (TPDB TRS_Standard/SK90/4.23)

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)
if(x,if(x,y,z),z) → if(x,y,z) (5)
if(x,y,if(x,y,z)) → if(x,y,z) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(false) = 0 status(false) = [] list-extension(false) = Lex
prec(if) = 0 status(if) = [1, 2, 3] list-extension(if) = Lex
prec(true) = 0 status(true) = [] list-extension(true) = Lex
and the following Max-polynomial interpretation
[false] = max(0)
[if(x1, x2, x3)] = max(0, 0 + 1 · x1, 0 + 1 · x2, 0 + 1 · x3)
[true] = max(4)
all of the following rules can be deleted.
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)
if(x,if(x,y,z),z) → if(x,y,z) (5)
if(x,y,if(x,y,z)) → if(x,y,z) (6)

1.1 R is empty

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