Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x01)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  f#(x1)	→	e#(x1)	(12)
  e#(b(x1))	→	c#(x1)	(9)
  c#(b(x1))	→	d#(x1)	(8)
  d#(b(x1))	→	f#(x1)	(11)
  e#(b(x1))	→	c#(c(x1))	(10)

  1.1.1.1 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

  [e(x1)]	=	-∞ 0 -∞ -∞ 0 0 -∞ 0 0 · x1 + 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞
  [d#(x1)]	=	-∞ 0 1 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞
  [b(x1)]	=	0 0 1 0 0 0 -∞ 0 0 · x1 + 0 -∞ -∞ 0 -∞ -∞ 0 -∞ -∞
  [d(x1)]	=	-∞ 0 1 0 -∞ 1 0 0 1 · x1 + 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞
  [f(x1)]	=	0 1 1 0 1 1 -∞ 0 0 · x1 + 1 -∞ -∞ 0 -∞ -∞ 0 -∞ -∞
  [f#(x1)]	=	0 1 1 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 1 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞
  [a(x1)]	=	0 1 1 0 0 1 0 0 0 · x1 + 1 -∞ -∞ 0 -∞ -∞ 0 -∞ -∞
  [c#(x1)]	=	0 0 0 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞
  [e#(x1)]	=	-∞ 0 0 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞
  [c(x1)]	=	0 0 0 0 0 0 0 0 0 · x1 + 0 -∞ -∞ 0 -∞ -∞ 0 -∞ -∞

  together with the usable rules

  a(x1)	→	b(b(x1))	(1)
  c(b(x1))	→	d(x1)	(2)
  e(b(x1))	→	c(c(x1))	(3)
  d(b(x1))	→	b(f(x1))	(4)
  f(x1)	→	a(e(x1))	(5)
  c(x1)	→	x1	(6)

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

  f#(x1)

  →

  e#(x1)

  (12)

  could be deleted.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.