Certification Problem

Input (TPDB TRS_Standard/Applicative_05/TreeFlatten)

The rewrite relation of the following TRS is considered.

app(app(map,f),nil) nil (1)
app(app(map,f),app(app(cons,x),xs)) app(app(cons,app(f,x)),app(app(map,f),xs)) (2)
app(flatten,app(app(node,x),xs)) app(app(cons,x),app(concat,app(app(map,flatten),xs))) (3)
app(concat,nil) nil (4)
app(concat,app(app(cons,x),xs)) app(app(append,x),app(concat,xs)) (5)
app(app(append,nil),xs) xs (6)
app(app(append,app(app(cons,x),xs)),ys) app(app(cons,x),app(app(append,xs),ys)) (7)

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(map,f),app(app(cons,x),xs)) app#(app(map,f),xs) (8)
app#(app(map,f),app(app(cons,x),xs)) app#(f,x) (9)
app#(app(map,f),app(app(cons,x),xs)) app#(cons,app(f,x)) (10)
app#(app(map,f),app(app(cons,x),xs)) app#(app(cons,app(f,x)),app(app(map,f),xs)) (11)
app#(flatten,app(app(node,x),xs)) app#(map,flatten) (12)
app#(flatten,app(app(node,x),xs)) app#(app(map,flatten),xs) (13)
app#(flatten,app(app(node,x),xs)) app#(concat,app(app(map,flatten),xs)) (14)
app#(flatten,app(app(node,x),xs)) app#(cons,x) (15)
app#(flatten,app(app(node,x),xs)) app#(app(cons,x),app(concat,app(app(map,flatten),xs))) (16)
app#(concat,app(app(cons,x),xs)) app#(concat,xs) (17)
app#(concat,app(app(cons,x),xs)) app#(append,x) (18)
app#(concat,app(app(cons,x),xs)) app#(app(append,x),app(concat,xs)) (19)
app#(app(append,app(app(cons,x),xs)),ys) app#(append,xs) (20)
app#(app(append,app(app(cons,x),xs)),ys) app#(app(append,xs),ys) (21)
app#(app(append,app(app(cons,x),xs)),ys) app#(app(cons,x),app(app(append,xs),ys)) (22)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.