Certification Problem

Input (TPDB TRS_Standard/AG01/#3.39)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ 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

  quot#(s(x),s(y))

  →

  quot#(minus(x,y),s(y))

  (10)

  1.1.1 Subterm Criterion Processor

  We use the projection to multisets

  π(quot#)	=	{ 1 }
  π(minus)	=	{ 1 }

  to remove the pairs:

  quot#(s(x),s(y))

  →

  quot#(minus(x,y),s(y))

  (10)

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  minus#(s(x),s(y))

  →

  minus#(x,y)

  (8)

  1.1.2 Size-Change Termination

  Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

  minus#(s(x),s(y))	→	minus#(x,y)	(8)
  2	>	2
  1	>	1

  As there is no critical graph in the transitive closure, there are no infinite chains.

• The 3^rd component contains the pair

  plus#(s(x),y)	→	plus#(x,y)	(11)
  plus#(minus(x,s(0)),minus(y,s(s(z))))	→	plus#(minus(y,s(s(z))),minus(x,s(0)))	(12)

  1.1.3 Subterm Criterion Processor

  We use the projection to multisets

  π(plus#)

  =

  { 1, 1, 2, 2 }

  to remove the pairs:

  plus#(s(x),y)

  →

  plus#(x,y)

  (11)

  1.1.3.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals

  [minus(x1, x2)]	=	1 1 0 1 1 1 0 1 0 0 1 1 1 0 0 1 · x1 + 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 1 · x2 + 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  [0]	=	0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
  [plus#(x1, x2)]	=	1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 · x1 + 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 · x2 + 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  [s(x1)]	=	0 1 0 0 0 1 0 0 0 0 0 0 1 0 1 1 · x1 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

  together with the usable rules

  minus(s(x),s(y))	→	minus(x,y)	(2)
  minus(x,0)	→	x	(1)

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

  plus#(minus(x,s(0)),minus(y,s(s(z))))

  →

  plus#(minus(y,s(s(z))),minus(x,s(0)))

  (12)

  could be deleted.

  1.1.3.1.1 P is empty

  There are no pairs anymore.