Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/05)

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#(b(b(x1)))	→	a#(a(a(x1)))	(9)
  a#(a(b(a(x1))))	→	b#(b(b(x1)))	(5)
  b#(b(b(x1)))	→	a#(a(x1))	(8)
  a#(a(b(a(x1))))	→	b#(b(x1))	(4)
  b#(b(b(x1)))	→	a#(x1)	(7)
  a#(a(b(a(x1))))	→	b#(x1)	(3)

  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 arctic semiring over the integers

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

  together with the usable rules

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

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

  b#(b(b(x1)))	→	a#(a(a(x1)))	(9)
  b#(b(b(x1)))	→	a#(a(x1))	(8)
  b#(b(b(x1)))	→	a#(x1)	(7)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.