Certification Problem

Input (TPDB TRS_Standard/Applicative_05/TreeHeight)

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(app(le,0),y) true (3)
app(app(le,app(s,x)),0) false (4)
app(app(le,app(s,x)),app(s,y)) app(app(le,x),y) (5)
app(app(maxlist,x),app(app(cons,y),ys)) app(app(if,app(app(le,x),y)),app(app(maxlist,y),ys)) (6)
app(app(maxlist,x),nil) x (7)
app(height,app(app(node,x),xs)) app(s,app(app(maxlist,0),app(app(map,height),xs))) (8)

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(map,f),app(app(cons,x),xs)) app#(cons,app(f,x)) (9)
app#(app(maxlist,x),app(app(cons,y),ys)) app#(if,app(app(le,x),y)) (10)
app#(height,app(app(node,x),xs)) app#(s,app(app(maxlist,0),app(app(map,height),xs))) (11)
app#(app(maxlist,x),app(app(cons,y),ys)) app#(app(maxlist,y),ys) (12)
app#(app(le,app(s,x)),app(s,y)) app#(app(le,x),y) (13)
app#(app(le,app(s,x)),app(s,y)) app#(le,x) (14)
app#(height,app(app(node,x),xs)) app#(app(maxlist,0),app(app(map,height),xs)) (15)
app#(app(maxlist,x),app(app(cons,y),ys)) app#(app(le,x),y) (16)
app#(app(maxlist,x),app(app(cons,y),ys)) app#(maxlist,y) (17)
app#(app(maxlist,x),app(app(cons,y),ys)) app#(app(if,app(app(le,x),y)),app(app(maxlist,y),ys)) (18)
app#(height,app(app(node,x),xs)) app#(map,height) (19)
app#(height,app(app(node,x),xs)) app#(app(map,height),xs) (20)
app#(app(map,f),app(app(cons,x),xs)) app#(f,x) (21)
app#(app(map,f),app(app(cons,x),xs)) app#(app(cons,app(f,x)),app(app(map,f),xs)) (22)
app#(app(maxlist,x),app(app(cons,y),ys)) app#(le,x) (23)
app#(height,app(app(node,x),xs)) app#(maxlist,0) (24)
app#(app(map,f),app(app(cons,x),xs)) app#(app(map,f),xs) (25)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.