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 AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f#(s(x)) f#(x) (8)
g#(s(x),s(y)) if#(f(x),s(x),s(y)) (9)
g#(s(x),s(y)) f#(x) (10)
g#(x,c(y)) g#(x,g(s(c(y)),y)) (11)
g#(x,c(y)) g#(s(c(y)),y) (12)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.