Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/013)

The rewrite relation of the following TRS is considered.

app(app(append,nil),ys) ys (1)
app(app(append,app(app(cons,x),xs)),ys) app(app(cons,x),app(app(append,xs),ys)) (2)
app(app(flatwith,f),app(leaf,x)) app(app(cons,app(f,x)),nil) (3)
app(app(flatwith,f),app(node,xs)) app(app(flatwithsub,f),xs) (4)
app(app(flatwithsub,f),nil) nil (5)
app(app(flatwithsub,f),app(app(cons,x),xs)) app(app(append,app(app(flatwith,f),x)),app(app(flatwithsub,f),xs)) (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,x),xs)),ys) app#(append,xs) (7)
app#(app(append,app(app(cons,x),xs)),ys) app#(app(append,xs),ys) (8)
app#(app(append,app(app(cons,x),xs)),ys) app#(app(cons,x),app(app(append,xs),ys)) (9)
app#(app(flatwith,f),app(leaf,x)) app#(f,x) (10)
app#(app(flatwith,f),app(leaf,x)) app#(cons,app(f,x)) (11)
app#(app(flatwith,f),app(leaf,x)) app#(app(cons,app(f,x)),nil) (12)
app#(app(flatwith,f),app(node,xs)) app#(flatwithsub,f) (13)
app#(app(flatwith,f),app(node,xs)) app#(app(flatwithsub,f),xs) (14)
app#(app(flatwithsub,f),app(app(cons,x),xs)) app#(app(flatwithsub,f),xs) (15)
app#(app(flatwithsub,f),app(app(cons,x),xs)) app#(flatwith,f) (16)
app#(app(flatwithsub,f),app(app(cons,x),xs)) app#(app(flatwith,f),x) (17)
app#(app(flatwithsub,f),app(app(cons,x),xs)) app#(append,app(app(flatwith,f),x)) (18)
app#(app(flatwithsub,f),app(app(cons,x),xs)) app#(app(append,app(app(flatwith,f),x)),app(app(flatwithsub,f),xs)) (19)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.