Certification Problem

Input (TPDB TRS_Standard/Der95/20)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  not#(or(x,y))	→	not#(not(not(y)))	(15)
  not#(and(x,y))	→	not#(y)	(8)
  not#(and(x,y))	→	not#(not(x))	(14)
  not#(or(x,y))	→	not#(not(not(x)))	(13)
  not#(or(x,y))	→	not#(x)	(12)
  not#(and(x,y))	→	not#(not(y))	(7)
  not#(and(x,y))	→	not#(x)	(6)
  not#(or(x,y))	→	not#(y)	(5)
  not#(or(x,y))	→	not#(not(y))	(11)
  not#(and(x,y))	→	not#(not(not(y)))	(10)
  not#(and(x,y))	→	not#(not(not(x)))	(9)
  not#(or(x,y))	→	not#(not(x))	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [and(x1, x2)]	=	x1 + x2 + 1
  [not#(x1)]	=	x1 + 0
  [or(x1, x2)]	=	x1 + x2 + 1
  [not(x1)]	=	x1 + 0

  together with the usable rules

  not(not(x))	→	x	(1)
  not(and(x,y))	→	or(not(not(not(x))),not(not(not(y))))	(3)
  not(or(x,y))	→	and(not(not(not(x))),not(not(not(y))))	(2)

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

  not#(or(x,y))	→	not#(not(not(y)))	(15)
  not#(and(x,y))	→	not#(y)	(8)
  not#(and(x,y))	→	not#(not(x))	(14)
  not#(or(x,y))	→	not#(not(not(x)))	(13)
  not#(or(x,y))	→	not#(x)	(12)
  not#(and(x,y))	→	not#(not(y))	(7)
  not#(and(x,y))	→	not#(x)	(6)
  not#(or(x,y))	→	not#(y)	(5)
  not#(or(x,y))	→	not#(not(y))	(11)
  not#(and(x,y))	→	not#(not(not(y)))	(10)
  not#(and(x,y))	→	not#(not(not(x)))	(9)
  not#(or(x,y))	→	not#(not(x))	(4)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.