Certification Problem

Input (TPDB SRS_Standard/Trafo_06/un08)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ 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

  b#(a(b(b(x1))))	→	b#(b(b(x1)))	(8)
  b#(b(b(x1)))	→	b#(a(b(x1)))	(9)
  b#(b(a(b(x1))))	→	b#(a(b(b(b(b(x1))))))	(7)
  b#(b(a(b(x1))))	→	b#(b(b(b(x1))))	(6)
  b#(b(a(b(x1))))	→	b#(b(b(x1)))	(5)
  b#(b(a(b(x1))))	→	b#(b(x1))	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

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

  together with the usable rules

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

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

  b#(b(a(b(x1))))	→	b#(a(b(b(b(b(x1))))))	(7)
  b#(b(a(b(x1))))	→	b#(b(b(b(x1))))	(6)
  b#(b(a(b(x1))))	→	b#(b(b(x1)))	(5)
  b#(b(a(b(x1))))	→	b#(b(x1))	(4)

  could be deleted.

  1.1.1.1 Subterm Criterion Processor

  We use the projection to multisets

  π(b#)	=	{ 1 }
  π(a)	=	{ 1 }

  to remove the pairs:

  b#(b(b(x1)))

  →

  b#(a(b(x1)))

  (9)

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.