Certification Problem

Input (TPDB TRS_Standard/AProVE_04/Liveness_WRS)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.

• The 1^st component contains the pair

  tp#(mark(x))

  →

  tp#(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x)))

  (31)

  1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [mat(x1, x2)]	=	1 · x1 + 1 · x2
  [f(x1)]	=	1 · x1
  [y]	=	0
  [c]	=	0
  [no(x1)]	=	1 · x1
  [chk(x1)]	=	1 · x1
  [X]	=	0
  [active(x1)]	=	1 · x1
  [mark(x1)]	=	1 · x1
  [tp#(x1)]	=	1 · x1

  together with the usable rules

  mat(f(x),f(y))	→	f(mat(x,y))	(3)
  mat(f(x),c)	→	no(c)	(5)
  chk(no(f(x)))	→	f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x)))	(2)
  chk(no(c))	→	active(c)	(4)
  f(active(x))	→	active(f(x))	(6)
  f(no(x))	→	no(f(x))	(7)
  f(mark(x))	→	mark(f(x))	(8)
  active(f(x))	→	mark(f(f(x)))	(1)

  (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.1.1 Reduction Pair Processor

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

  [tp#(x1)]	=	-∞ + 0 · x1
  [mark(x1)]	=	3 + 1 · x1
  [chk(x1)]	=	-∞ + 0 · x1
  [mat(x1, x2)]	=	2 + -∞ · x1 + 0 · x2
  [f(x1)]	=	1 + 1 · x1
  [X]	=	0
  [y]	=	2
  [c]	=	0
  [no(x1)]	=	2 + 2 · x1
  [active(x1)]	=	2 + 2 · x1

  the pair

  tp#(mark(x))

  →

  tp#(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x)))

  (31)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  chk#(no(f(x)))

  →

  chk#(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x))

  (12)

  1.1.2 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [mat(x1, x2)]	=	1 · x1 + 1 · x2
  [f(x1)]	=	1 · x1
  [y]	=	0
  [c]	=	0
  [no(x1)]	=	1 · x1
  [X]	=	0
  [chk#(x1)]	=	1 · x1

  together with the usable rules

  mat(f(x),f(y))	→	f(mat(x,y))	(3)
  mat(f(x),c)	→	no(c)	(5)

  (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.2.1 Switch to Innermost Termination

  The TRS does not have overlaps with the pairs and is locally confluent:

  20

  Hence, it suffices to show innermost termination in the following.

  1.1.2.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [mat(x1, x2)]	=	1 · x1 + 2 · x2
  [f(x1)]	=	2 · x1
  [y]	=	0
  [c]	=	1
  [no(x1)]	=	1 · x1
  [chk#(x1)]	=	2 · x1
  [X]	=	0

  the rule

  mat(f(x),c)

  →

  no(c)

  (5)

  could be deleted.

  1.1.2.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [mat(x1, x2)]	=	2 · x1 + 2 · x2
  [f(x1)]	=	2 · x1
  [y]	=	0
  [chk#(x1)]	=	1 · x1
  [no(x1)]	=	2 · x1
  [X]	=	0

  together with the usable rule

  mat(f(x),f(y))

  →

  f(mat(x,y))

  (3)

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

  chk#(no(f(x)))

  →

  chk#(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))),x))

  (12)

  and no rules could be deleted.

  1.1.2.1.1.1.1 P is empty

  There are no pairs anymore.

• The 3^rd component contains the pair

  f#(active(x))	→	active#(f(x))	(27)
  active#(f(x))	→	f#(f(x))	(10)
  f#(active(x))	→	f#(x)	(28)
  f#(no(x))	→	f#(x)	(29)
  f#(mark(x))	→	f#(x)	(30)

  1.1.3 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [f(x1)]	=	1 · x1
  [active(x1)]	=	1 · x1
  [no(x1)]	=	1 · x1
  [mark(x1)]	=	1 · x1
  [active#(x1)]	=	1 · x1
  [f#(x1)]	=	1 · x1

  together with the usable rules

  f(active(x))	→	active(f(x))	(6)
  f(no(x))	→	no(f(x))	(7)
  f(mark(x))	→	mark(f(x))	(8)
  active(f(x))	→	mark(f(f(x)))	(1)

  (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.3.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [f(x1)]	=	2 · x1
  [active(x1)]	=	2 + 2 · x1
  [no(x1)]	=	1 + 2 · x1
  [mark(x1)]	=	1 + 1 · x1
  [f#(x1)]	=	1 + 2 · x1
  [active#(x1)]	=	2 + 2 · x1

  the pairs

  f#(active(x))	→	active#(f(x))	(27)
  active#(f(x))	→	f#(f(x))	(10)
  f#(active(x))	→	f#(x)	(28)
  f#(no(x))	→	f#(x)	(29)
  f#(mark(x))	→	f#(x)	(30)

  and the rules

  f(active(x))	→	active(f(x))	(6)
  f(no(x))	→	no(f(x))	(7)
  f(mark(x))	→	mark(f(x))	(8)
  active(f(x))	→	mark(f(f(x)))	(1)

  could be deleted.

  1.1.3.1.1 P is empty

  There are no pairs anymore.