Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/015)

The rewrite relation of the following TRS is considered.

app(app(app(if,true),x),y) x (1)
app(app(app(if,false),x),y) y (2)
app(app(filter,f),nil) nil (3)
app(app(filter,f),app(app(cons,x),xs)) app(app(app(if,app(f,x)),app(app(cons,x),app(app(filter,f),xs))),app(app(filter,f),xs)) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
app#(app(filter,f),app(app(cons,x),xs)) app#(app(filter,f),xs) (5)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(cons,x),app(app(filter,f),xs)) (6)
app#(app(filter,f),app(app(cons,x),xs)) app#(f,x) (7)
app#(app(filter,f),app(app(cons,x),xs)) app#(if,app(f,x)) (8)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(if,app(f,x)),app(app(cons,x),app(app(filter,f),xs))) (9)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(if,app(f,x)),app(app(cons,x),app(app(filter,f),xs))),app(app(filter,f),xs)) (10)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.