Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-126)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

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

  c#(b(x1))	→	b#(a(x1))	(11)
  b#(a(a(x1)))	→	c#(x1)	(6)
  b#(a(a(x1)))	→	c#(c(x1))	(7)
  b#(a(a(x1)))	→	b#(c(c(x1)))	(8)

  1.1.1.1 Subterm Criterion Processor

  We use the projection to multisets

  π(c#)	=	{ 1, 1 }
  π(b#)	=	{ 1, 1 }
  π(c)	=	{ 1 }
  π(b)	=	{ 1, 1 }
  π(a)	=	{ 1 }

  to remove the pairs:

  c#(b(x1))

  →

  b#(a(x1))

  (11)

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    b#(a(a(x1)))

    →

    b#(c(c(x1)))

    (8)

    1.1.1.1.1.1 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

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

    together with the usable rules

    a(x1)	→	x1	(1)
    b(a(a(x1)))	→	a(b(c(c(x1))))	(4)
    c(b(x1))	→	b(a(x1))	(5)

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

    b#(a(a(x1)))

    →

    b#(c(c(x1)))

    (8)

    could be deleted.

    1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.