Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/021)

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(times,0),y) 0 (3)
app(app(times,app(s,x)),y) app(app(plus,app(app(times,x),y)),y) (4)
app(inc,xs) app(app(map,app(plus,app(s,0))),xs) (5)
app(double,xs) app(app(map,app(times,app(s,app(s,0)))),xs) (6)
app(app(map,f),nil) nil (7)
app(app(map,f),app(app(cons,x),xs)) app(app(cons,app(f,x)),app(app(map,f),xs)) (8)

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) (9)
app#(app(plus,app(s,x)),y) app#(app(plus,x),y) (10)
app#(app(plus,app(s,x)),y) app#(s,app(app(plus,x),y)) (11)
app#(app(times,app(s,x)),y) app#(times,x) (12)
app#(app(times,app(s,x)),y) app#(app(times,x),y) (13)
app#(app(times,app(s,x)),y) app#(plus,app(app(times,x),y)) (14)
app#(app(times,app(s,x)),y) app#(app(plus,app(app(times,x),y)),y) (15)
app#(inc,xs) app#(s,0) (16)
app#(inc,xs) app#(plus,app(s,0)) (17)
app#(inc,xs) app#(map,app(plus,app(s,0))) (18)
app#(inc,xs) app#(app(map,app(plus,app(s,0))),xs) (19)
app#(double,xs) app#(s,0) (20)
app#(double,xs) app#(s,app(s,0)) (21)
app#(double,xs) app#(times,app(s,app(s,0))) (22)
app#(double,xs) app#(map,app(times,app(s,app(s,0)))) (23)
app#(double,xs) app#(app(map,app(times,app(s,app(s,0)))),xs) (24)
app#(app(map,f),app(app(cons,x),xs)) app#(app(map,f),xs) (25)
app#(app(map,f),app(app(cons,x),xs)) app#(f,x) (26)
app#(app(map,f),app(app(cons,x),xs)) app#(cons,app(f,x)) (27)
app#(app(map,f),app(app(cons,x),xs)) app#(app(cons,app(f,x)),app(app(map,f),xs)) (28)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.