Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/aprove09)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

1.1 String Reversal

1.1.1 Dependency Pair Transformation

1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  0#(p(x1))

  →

  0#(x1)

  (13)

  1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1

  [p(x1)]	=	x1 + 1
  [0#(x1)]	=	x1 + 0

  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

  0#(p(x1))

  →

  0#(x1)

  (13)

  and no rules could be deleted.

  1.1.1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  s#(q(x1))

  →

  s#(s(p(p(r(s(s(s(s(s(s(p(p(x1)))))))))))))

  (27)

  1.1.1.1.1.2 Reduction Pair Processor with Usable Rules

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

  [s(x1)]	=	0 0 0 0 0 0 -3 0 0 · x1 + -∞ -∞ -∞
  [r(x1)]	=	0 0 0 0 0 0 -3 -3 -3 · x1 + -∞ -∞ -∞
  [q(x1)]	=	0 0 0 0 0 0 0 0 0 · x1 + -∞ -∞ -∞
  [p(x1)]	=	0 0 0 -3 -3 0 -3 -3 0 · x1 + -∞ -∞ -∞
  [s#(x1)]	=	1 1 3 1 1 3 1 1 3 · x1 + -∞ -∞ -∞

  together with the usable rules

  s(q(x1))	→	s(s(p(p(r(s(s(s(s(s(s(p(p(x1)))))))))))))	(8)
  s(r(x1))	→	s(p(s(p(q(s(s(p(s(p(x1))))))))))	(9)
  s(p(p(x1)))	→	p(x1)	(10)
  s(p(x1))	→	x1	(11)

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

  s#(q(x1))

  →

  s#(s(p(p(r(s(s(s(s(s(s(p(p(x1)))))))))))))

  (27)

  could be deleted.

  1.1.1.1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.