Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.28)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by NaTT @ 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

  bits#(s(x))

  →

  bits#(half(s(x)))

  (8)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the argument filter
  in combination with the following Weighted Path Order with the following precedence and status

  prec(s)	=	1		status(s)	=	[1]		list-extension(s)	=	Lex
  prec(half#)	=	0		status(half#)	=	[]		list-extension(half#)	=	Lex
  prec(half)	=	1		status(half)	=	[]		list-extension(half)	=	Lex
  prec(0)	=	0		status(0)	=	[]		list-extension(0)	=	Lex
  prec(bits#)	=	0		status(bits#)	=	[1]		list-extension(bits#)	=	Lex
  prec(bits)	=	0		status(bits)	=	[]		list-extension(bits)	=	Lex

  and the following Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [half#(x1)]	=	1
  [half(x1)]	=	x1 + 0
  [0]	=	20653
  [bits#(x1)]	=	x1 + 1
  [bits(x1)]	=	1

  together with the usable rules

  half(0)	→	0	(1)
  half(s(s(x)))	→	s(half(x))	(3)
  half(s(0))	→	0	(2)

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

  bits#(s(x))

  →

  bits#(half(s(x)))

  (8)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  half#(s(s(x)))

  →

  half#(x)

  (7)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [half#(x1)]	=	x1 + 0
  [half(x1)]	=	x1 + 41141
  [0]	=	20163
  [bits#(x1)]	=	0
  [bits(x1)]	=	0

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

  half#(s(s(x)))

  →

  half#(x)

  (7)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.