Certification Problem

Input (TPDB SRS_Standard/Wenzel_16/abcabbcab-abbcabcabbca.srs)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

{c(☐), b(☐), a(☐)}

1.1 Semantic Labeling

As carrier we take the set {0,1,2}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 3):

1.1.1 Rule Removal

1.1.1.1 Dependency Pair Transformation

1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  c2#(a1(b0(c2(a1(b1(b0(c2(a1(b1(x1))))))))))	→	c2#(a1(b0(c2(a1(b1(b0(c2(a1(x1)))))))))	(19)
  c2#(a1(b0(c2(a1(b1(b0(c2(a1(b1(x1))))))))))	→	c2#(a1(x1))	(18)

  1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

  [c2(x1)]	=	-∞ 0 -∞ -∞ -∞ 0 0 0 -∞ · x1 + 13 0 -∞
  [b0(x1)]	=	-∞ -∞ 0 -∞ 1 -∞ 0 -∞ -∞ · x1 + 3 -∞ 6
  [b1(x1)]	=	-∞ 0 -∞ 11 -∞ 2 0 -∞ 1 · x1 + 7 13 14
  [a0(x1)]	=	14 16 4 -∞ 0 1 -∞ -∞ -∞ · x1 + -∞ 16 0
  [a1(x1)]	=	-∞ -∞ -∞ 0 -∞ -∞ -∞ -∞ 0 · x1 + 3 0 13
  [c2#(x1)]	=	9 -∞ 1 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 2 -∞ -∞

  together with the usable rules

  c2(a1(b0(c2(a1(b1(b0(c2(a1(b1(x1))))))))))	→	c2(a1(b1(b0(c2(a1(b0(c2(a1(b1(b0(c2(a1(x1)))))))))))))	(12)
  c2(a1(b0(c2(a1(b1(b0(c2(a1(b0(x1))))))))))	→	c2(a1(b1(b0(c2(a1(b0(c2(a1(b1(b0(c2(a0(x1)))))))))))))	(13)

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

  c2#(a1(b0(c2(a1(b1(b0(c2(a1(b1(x1))))))))))	→	c2#(a1(b0(c2(a1(b1(b0(c2(a1(x1)))))))))	(19)
  c2#(a1(b0(c2(a1(b1(b0(c2(a1(b1(x1))))))))))	→	c2#(a1(x1))	(18)

  could be deleted.

  1.1.1.1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.