Certification Problem

Input (TPDB TRS_Standard/SK90/2.43)

The rewrite relation of the following TRS is considered.

merge(nil,y)	→	y	(1)
merge(x,nil)	→	x	(2)
merge(.(x,y),.(u,v))	→	if(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))	(3)
++(nil,y)	→	y	(4)
++(.(x,y),z)	→	.(x,++(y,z))	(5)
if(true,x,y)	→	x	(6)
if(false,x,y)	→	x	(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.

merge^#(.(x,y),.(u,v))	→	merge^#(.(x,y),v)	(8)
merge^#(.(x,y),.(u,v))	→	merge^#(y,.(u,v))	(9)
merge^#(.(x,y),.(u,v))	→	if^#(<(x,u),.(x,merge(y,.(u,v))),.(u,merge(.(x,y),v)))	(10)
++^#(.(x,y),z)	→	++^#(y,z)	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

merge^#(.(x,y),.(u,v)) → merge^#(.(x,y),v) (8)

merge^#(.(x,y),.(u,v)) → merge^#(y,.(u,v)) (9)

1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

merge^#(.(x,y),.(u,v)) → merge^#(.(x,y),v) (8)

2 > 2

1 ≥ 1

merge^#(.(x,y),.(u,v)) → merge^#(y,.(u,v)) (9)

2 ≥ 2

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.
The 2^nd component contains the pair
++^#(.(x,y),z) → ++^#(y,z) (11)

1.1.2 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

++^#(.(x,y),z) → ++^#(y,z) (11)

2 ≥ 2

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.