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 AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[if(x1, x2, x3)] = 2 · x1 + 2 · x2 + 2 · x3
[true] = 2
[false] = 2
[u] = 0
[v] = 0
all of the following rules can be deleted.
if(true,x,y) x (1)
if(false,x,y) y (2)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[if(x1, x2, x3)] = 1 + 2 · x1 + 1 · x2 + 1 · x3
[u] = 0
[v] = 0
all of the following rules can be deleted.
if(x,y,y) y (3)

1.1.1 Rule Removal

Using the
prec(if) = 0 stat(if) = lex
prec(u) = 1 stat(u) = lex
prec(v) = 1 stat(v) = lex

π(if) = [1,2,3]
π(u) = []
π(v) = []

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 R is empty

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