Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/028)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
app#(app(app(consif,true),x),ys) app#(cons,x) (5)
app#(app(filter,f),app(app(cons,x),xs)) app#(f,x) (6)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(filter,f),xs) (7)
app#(app(filter,f),app(app(cons,x),xs)) app#(consif,app(f,x)) (8)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(app(consif,app(f,x)),x),app(app(filter,f),xs)) (9)
app#(app(app(consif,true),x),ys) app#(app(cons,x),ys) (10)
app#(app(filter,f),app(app(cons,x),xs)) app#(app(consif,app(f,x)),x) (11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.