Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-246)

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 2 components.

• The 1^st component contains the pair

  a#(b(b(b(x1))))	→	a#(x1)	(12)
  a#(b(b(b(x1))))	→	a#(x1)	(12)
  a#(b(b(b(x1))))	→	a#(a(x1))	(7)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 7720
  [b(x1)]	=	x1 + 7720
  [a#(x1)]	=	x1 + 0
  [b#(x1)]	=	0

  together with the usable rules

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

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

  a#(b(b(b(x1))))	→	a#(x1)	(12)
  a#(b(b(b(x1))))	→	a#(x1)	(12)
  a#(b(b(b(x1))))	→	a#(a(x1))	(7)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  b#(a(a(a(x1))))	→	b#(b(a(a(x1))))	(6)
  b#(a(a(a(x1))))	→	b#(a(a(x1)))	(8)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 1
  [b(x1)]	=	2
  [a#(x1)]	=	x1 + 0
  [b#(x1)]	=	x1 + 0

  together with the usable rules

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

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

  b#(a(a(a(x1))))	→	b#(b(a(a(x1))))	(6)
  b#(a(a(a(x1))))	→	b#(a(a(x1)))	(8)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.