Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/6)

The rewrite relation of the following TRS is considered.

b(x,y) c(a(c(y),a(0,x))) (1)
a(y,x) y (2)
a(y,c(b(a(0,x),0))) b(a(c(b(0,y)),x),0) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(x,y) a#(c(y),a(0,x)) (4)
b#(x,y) a#(0,x) (5)
a#(y,c(b(a(0,x),0))) b#(a(c(b(0,y)),x),0) (6)
a#(y,c(b(a(0,x),0))) a#(c(b(0,y)),x) (7)
a#(y,c(b(a(0,x),0))) b#(0,y) (8)

1.1 Instantiation Processor

We instantiate the pair to the following set of pairs
b#(y_3,0) a#(c(0),a(0,y_3)) (9)
b#(0,y_0) a#(c(y_0),a(0,0)) (10)

1.1.1 Instantiation Processor

We instantiate the pair to the following set of pairs
b#(y_3,0) a#(0,y_3) (11)
b#(0,y_0) a#(0,0) (12)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.