Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/004)

The rewrite relation of the following TRS is considered.

app(app(plus,0),y) y (1)
app(app(plus,app(s,x)),y) app(s,app(app(plus,x),y)) (2)
app(app(sumwith,f),nil) nil (3)
app(app(sumwith,f),app(app(cons,x),xs)) app(app(plus,app(f,x)),app(app(sumwith,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(plus,app(s,x)),y) app#(plus,x) (5)
app#(app(plus,app(s,x)),y) app#(app(plus,x),y) (6)
app#(app(plus,app(s,x)),y) app#(s,app(app(plus,x),y)) (7)
app#(app(sumwith,f),app(app(cons,x),xs)) app#(app(sumwith,f),xs) (8)
app#(app(sumwith,f),app(app(cons,x),xs)) app#(f,x) (9)
app#(app(sumwith,f),app(app(cons,x),xs)) app#(plus,app(f,x)) (10)
app#(app(sumwith,f),app(app(cons,x),xs)) app#(app(plus,app(f,x)),app(app(sumwith,f),xs)) (11)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.