Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z065)

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

  ##(0(x1))	→	##(x1)	(14)
  1#(*(x1))	→	0#(#(x1))	(13)
  ##(0(x1))	→	0#(#(x1))	(12)
  ##(1(x1))	→	1#(#(x1))	(9)
  0#(*(x1))	→	1#(x1)	(11)
  ##(1(x1))	→	##(x1)	(10)
  1#(*(x1))	→	##(x1)	(8)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [0#(x1)]	=	x1 + 0
  [1(x1)]	=	x1 + 21240
  [##(x1)]	=	x1 + 15440
  [#(x1)]	=	x1 + 7721
  [0(x1)]	=	x1 + 21240
  [$(x1)]	=	1
  [1#(x1)]	=	x1 + 7720
  [*(x1)]	=	x1 + 7721

  together with the usable rules

  #(1(x1))	→	1(#(x1))	(4)
  0(*(x1))	→	*(1(x1))	(1)
  #(0(x1))	→	0(#(x1))	(3)
  #($(x1))	→	*($(x1))	(5)
  #(*(x1))	→	*(x1)	(7)
  #(#(x1))	→	#(x1)	(6)
  1(*(x1))	→	0(#(x1))	(2)

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

  ##(0(x1))	→	##(x1)	(14)
  1#(*(x1))	→	0#(#(x1))	(13)
  ##(0(x1))	→	0#(#(x1))	(12)
  ##(1(x1))	→	1#(#(x1))	(9)
  0#(*(x1))	→	1#(x1)	(11)
  ##(1(x1))	→	##(x1)	(10)
  1#(*(x1))	→	##(x1)	(8)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.