Certification Problem

Input (TPDB SRS_Standard/Waldmann_06_SRS/uni-5)

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

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

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers

  [b#(x1)]	=	-3 0 -∞ -∞ · x1 + 0 -∞ -∞ -∞
  [b(x1)]	=	-∞ 0 0 1 · x1 + 1 -∞ 2 -∞
  [c#(x1)]	=	0 0 -∞ -∞ · x1 + 0 -∞ -∞ -∞
  [a(x1)]	=	-∞ 3 -3 0 · x1 + 0 -∞ 0 -∞
  [c(x1)]	=	0 1 -2 -∞ · x1 + 2 -∞ -2 -∞

  together with the usable rules

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

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

  c#(c(x1))

  →

  b#(x1)

  (8)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    c#(c(x1))

    →

    c#(b(x1))

    (9)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals

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

    together with the usable rules

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

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

    c#(c(x1))

    →

    c#(b(x1))

    (9)

    could be deleted.

    1.1.1.1.1.1 P is empty

    There are no pairs anymore.