Certification Problem

Input (TPDB TRS_Standard/SK90/2.44)

The rewrite relation of the following TRS is considered.

del(.(x,.(y,z)))	→	f(=(x,y),x,y,z)	(1)
f(true,x,y,z)	→	del(.(y,z))	(2)
f(false,x,y,z)	→	.(x,del(.(y,z)))	(3)
=(nil,nil)	→	true	(4)
=(.(x,y),nil)	→	false	(5)
=(nil,.(y,z))	→	false	(6)
=(.(x,y),.(u,v))	→	and(=(x,u),=(y,v))	(7)

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.

=^#(.(x,y),.(u,v))	→	=^#(x,u)	(8)
del^#(.(x,.(y,z)))	→	=^#(x,y)	(9)
del^#(.(x,.(y,z)))	→	f^#(=(x,y),x,y,z)	(10)
=^#(.(x,y),.(u,v))	→	=^#(y,v)	(11)
f^#(true,x,y,z)	→	del^#(.(y,z))	(12)
f^#(false,x,y,z)	→	del^#(.(y,z))	(13)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(false,x,y,z)	→	del^#(.(y,z))	(13)
f^#(true,x,y,z)	→	del^#(.(y,z))	(12)
del^#(.(x,.(y,z)))	→	f^#(=(x,y),x,y,z)	(10)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[del^#(x₁)]	=	x₁ + 0
[v]	=	1
[u]	=	1
[and(x₁, x₂)]	=	x₁ + 0
[false]	=	1
[true]	=	1
[f(x₁,...,x₄)]	=	0
[del(x₁)]	=	0
[=(x₁, x₂)]	=	1
[nil]	=	1
[f^#(x₁,...,x₄)]	=	x₁ + x₂ + x₃ + x₄ + 2
[.(x₁, x₂)]	=	x₁ + x₂ + 2
[=^#(x₁, x₂)]	=	0

together with the usable rules

=(nil,nil)	→	true	(4)
=(.(x,y),nil)	→	false	(5)
=(.(x,y),.(u,v))	→	and(=(x,u),=(y,v))	(7)
=(nil,.(y,z))	→	false	(6)

(w.r.t. the implicit argument filter of the reduction pair), the pairs

f^#(false,x,y,z)	→	del^#(.(y,z))	(13)
f^#(true,x,y,z)	→	del^#(.(y,z))	(12)
del^#(.(x,.(y,z)))	→	f^#(=(x,y),x,y,z)	(10)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.