Certification Problem

Input (TPDB TRS_Standard/AProVE_04/rta1)

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 2 components.

• The 1^st component contains the pair

  ack#(s(x),0)	→	ack#(x,s(0))	(10)
  ack#(s(x),s(y))	→	ack#(x,plus(y,ack(s(x),y)))	(13)
  ack#(s(x),s(y))	→	ack#(s(x),y)	(11)

  1.1.1 Size-Change Termination

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

  ack#(s(x),0)	→	ack#(x,s(0))	(10)
  1	>	1
  ack#(s(x),s(y))	→	ack#(x,plus(y,ack(s(x),y)))	(13)
  1	>	1
  ack#(s(x),s(y))	→	ack#(s(x),y)	(11)
  2	>	2
  1	≥	1

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

• The 2^nd component contains the pair

  plus#(x,s(s(y)))	→	plus#(s(x),y)	(9)
  plus#(s(s(x)),y)	→	plus#(x,s(y))	(8)

  1.1.2 Reduction Pair Processor with Usable Rules

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

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

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

  plus#(x,s(s(y)))	→	plus#(s(x),y)	(9)
  plus#(s(s(x)),y)	→	plus#(x,s(y))	(8)

  could be deleted.

  1.1.2.1 P is empty

  There are no pairs anymore.