Certification Problem

Input (TPDB TRS_Standard/Applicative_05/Ex3Lists)

The rewrite relation of the following TRS is considered.

app(app(append,nil),l) l (1)
app(app(append,app(app(cons,h),t)),l) app(app(cons,h),app(app(append,t),l)) (2)
app(app(map,f),nil) nil (3)
app(app(map,f),app(app(cons,h),t)) app(app(cons,app(f,h)),app(app(map,f),t)) (4)
app(app(append,app(app(append,l1),l2)),l3) app(app(append,l1),app(app(append,l2),l3)) (5)
app(app(map,f),app(app(append,l1),l2)) app(app(append,app(app(map,f),l1)),app(app(map,f),l2)) (6)

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(append,app(app(cons,h),t)),l) app#(append,t) (7)
app#(app(append,app(app(cons,h),t)),l) app#(app(append,t),l) (8)
app#(app(append,app(app(cons,h),t)),l) app#(app(cons,h),app(app(append,t),l)) (9)
app#(app(map,f),app(app(cons,h),t)) app#(app(map,f),t) (10)
app#(app(map,f),app(app(cons,h),t)) app#(f,h) (11)
app#(app(map,f),app(app(cons,h),t)) app#(cons,app(f,h)) (12)
app#(app(map,f),app(app(cons,h),t)) app#(app(cons,app(f,h)),app(app(map,f),t)) (13)
app#(app(append,app(app(append,l1),l2)),l3) app#(append,l2) (14)
app#(app(append,app(app(append,l1),l2)),l3) app#(app(append,l2),l3) (15)
app#(app(append,app(app(append,l1),l2)),l3) app#(app(append,l1),app(app(append,l2),l3)) (16)
app#(app(map,f),app(app(append,l1),l2)) app#(app(map,f),l2) (17)
app#(app(map,f),app(app(append,l1),l2)) app#(app(map,f),l1) (18)
app#(app(map,f),app(app(append,l1),l2)) app#(append,app(app(map,f),l1)) (19)
app#(app(map,f),app(app(append,l1),l2)) app#(app(append,app(app(map,f),l1)),app(app(map,f),l2)) (20)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.