Certification Problem

Input (TPDB TRS_Standard/AG01/#3.19)

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

  plus#(s(plus(x,y)),z)	→	plus#(plus(x,y),z)	(17)
  plus#(s(x),y)	→	plus#(x,y)	(12)
  plus#(s(x),y)	→	plus#(x,s(y))	(13)
  plus#(s(x),y)	→	plus#(minus(x,y),double(y))	(16)

  1.1.1 Subterm Criterion Processor

  We use the projection to multisets

  π(plus#)	=	{ 1 }
  π(minus)	=	{ 1 }

  to remove the pairs:

  plus#(s(plus(x,y)),z)	→	plus#(plus(x,y),z)	(17)
  plus#(s(x),y)	→	plus#(x,y)	(12)
  plus#(s(x),y)	→	plus#(x,s(y))	(13)
  plus#(s(x),y)	→	plus#(minus(x,y),double(y))	(16)

  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)

  (10)

  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)	(10)
  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

  double#(s(x))

  →

  double#(x)

  (11)

  1.1.3 Size-Change Termination

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

  double#(s(x))	→	double#(x)	(11)
  1	>	1

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