Certification Problem

Input (TPDB SRS_Standard/Trafo_06/hom02)

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

  a#(b(b(x1)))	→	c#(c(a(x1)))	(16)
  c#(c(c(b(b(x1)))))	→	a#(x1)	(9)
  b#(b(x1))	→	c#(c(x1))	(15)
  a#(b(b(x1)))	→	b#(b(c(c(c(a(x1))))))	(14)
  a#(b(b(x1)))	→	c#(c(c(a(x1))))	(13)
  a#(x1)	→	b#(x1)	(8)
  b#(b(x1))	→	c#(x1)	(7)
  a#(b(b(x1)))	→	a#(x1)	(6)
  a#(b(b(x1)))	→	c#(a(x1))	(12)
  a#(x1)	→	b#(b(x1))	(11)
  b#(b(x1))	→	c#(c(c(x1)))	(10)
  a#(b(b(x1)))	→	b#(c(c(c(a(x1)))))	(5)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 8
  [b(x1)]	=	x1 + 4
  [c(x1)]	=	x1 + 0
  [c#(x1)]	=	x1 + 0
  [a#(x1)]	=	x1 + 6
  [b#(x1)]	=	x1 + 0

  together with the usable rules

  c(c(c(b(b(x1)))))	→	a(x1)	(4)
  a(x1)	→	b(b(x1))	(1)
  b(b(x1))	→	c(c(c(x1)))	(3)
  a(b(b(x1)))	→	b(b(c(c(c(a(x1))))))	(2)

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

  a#(b(b(x1)))	→	c#(c(a(x1)))	(16)
  c#(c(c(b(b(x1)))))	→	a#(x1)	(9)
  b#(b(x1))	→	c#(c(x1))	(15)
  a#(b(b(x1)))	→	b#(b(c(c(c(a(x1))))))	(14)
  a#(b(b(x1)))	→	c#(c(c(a(x1))))	(13)
  a#(x1)	→	b#(x1)	(8)
  b#(b(x1))	→	c#(x1)	(7)
  a#(b(b(x1)))	→	a#(x1)	(6)
  a#(b(b(x1)))	→	c#(a(x1))	(12)
  a#(x1)	→	b#(b(x1))	(11)
  b#(b(x1))	→	c#(c(c(x1)))	(10)
  a#(b(b(x1)))	→	b#(c(c(c(a(x1)))))	(5)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.