Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/matchbox2)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Reduction Pair Processor

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  a#(l(x1))

  →

  a#(t(x1))

  (10)

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [a#(x1)]	=	x1
  [t(x1)]	=	2 · x1
  [o(x1)]	=	1 + x1
  [m(x1)]	=	-2 + 2 · x1
  [a(x1)]	=	2
  [e(x1)]	=	1
  [n(x1)]	=	2
  [s(x1)]	=	0
  [l(x1)]	=	2 + 2 · x1

  together with the usable rules

  t(o(x1))	→	m(a(x1))	(1)
  t(e(x1))	→	n(s(x1))	(2)
  a(l(x1))	→	a(t(x1))	(3)
  n(s(x1))	→	a(l(a(t(x1))))	(6)

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

  a#(l(x1))

  →

  a#(t(x1))

  (10)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  t#(e(x1))	→	s#(x1)	(9)
  s#(a(x1))	→	o#(m(a(t(e(x1)))))	(16)
  o#(m(a(x1)))	→	t#(e(n(x1)))	(12)
  s#(a(x1))	→	t#(e(x1))	(18)

  1.1.1.2 Reduction Pair Processor

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

  [t#(x1)]	=	-∞ -∞ -∞ + 0 0 1 -∞ -∞ -∞ -∞ -∞ -∞ · x1
  [e(x1)]	=	-∞ -∞ -∞ + 0 0 0 -∞ 0 -∞ -∞ -∞ 0 · x1
  [s#(x1)]	=	-∞ -∞ -∞ + 0 0 1 -∞ -∞ -∞ -∞ -∞ -∞ · x1
  [a(x1)]	=	-∞ -∞ -∞ + 0 -∞ -∞ -∞ -∞ -∞ 0 1 0 · x1
  [o#(x1)]	=	-∞ -∞ -∞ + 0 -∞ 0 -∞ -∞ -∞ -∞ -∞ -∞ · x1
  [m(x1)]	=	-∞ 0 -∞ + 0 0 -∞ 0 0 0 -∞ 0 0 · x1
  [t(x1)]	=	-∞ -∞ -∞ + 0 -∞ -∞ -∞ 0 -∞ 0 0 -∞ · x1
  [n(x1)]	=	-∞ -∞ -∞ + 0 0 0 -∞ -∞ 0 -∞ 0 -∞ · x1
  [s(x1)]	=	-∞ -∞ -∞ + 0 0 0 0 0 0 -∞ 0 -∞ · x1
  [l(x1)]	=	-∞ -∞ -∞ + 0 0 -∞ -∞ 0 -∞ -∞ 0 -∞ · x1
  [o(x1)]	=	-∞ 0 0 + 0 -∞ 0 0 1 0 1 0 0 · x1

  the pair

  s#(a(x1))

  →

  o#(m(a(t(e(x1)))))

  (16)

  could be deleted.

  1.1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    s#(a(x1))	→	t#(e(x1))	(18)
    t#(e(x1))	→	s#(x1)	(9)

    1.1.1.2.1.1 Monotonic Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

    [a(x1)]	=	1 · x1
    [e(x1)]	=	1 · x1
    [t#(x1)]	=	1 · x1
    [s#(x1)]	=	1 · x1

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

    1.1.1.2.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

    [s#(x1)]	=	1 · x1
    [a(x1)]	=	1 + 1 · x1
    [t#(x1)]	=	1 · x1
    [e(x1)]	=	1 + 1 · x1

    the pair

    t#(e(x1))

    →

    s#(x1)

    (9)

    could be deleted.

    1.1.1.2.1.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

    [s#(x1)]	=	1 + 1 · x1
    [a(x1)]	=	1 + 1 · x1
    [t#(x1)]	=	1 + 1 · x1
    [e(x1)]	=	1 · x1

    the pair

    s#(a(x1))

    →

    t#(e(x1))

    (18)

    could be deleted.

    1.1.1.2.1.1.1.1.1 P is empty

    There are no pairs anymore.