Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.27)

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

  minus#(x,s(y))

  →

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

  (15)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the argument filter

  π(le)	=	2
  π(if)	=	3
  π(minus#)	=	2

  in combination with the following Weighted Path Order with the following precedence and status

  prec(s)	=	0		status(s)	=	[1]		list-extension(s)	=	Lex
  prec(le#)	=	0		status(le#)	=	[2]		list-extension(le#)	=	Lex
  prec(minus)	=	0		status(minus)	=	[]		list-extension(minus)	=	Lex
  prec(false)	=	0		status(false)	=	[]		list-extension(false)	=	Lex
  prec(p#)	=	0		status(p#)	=	[]		list-extension(p#)	=	Lex
  prec(true)	=	0		status(true)	=	[]		list-extension(true)	=	Lex
  prec(p)	=	0		status(p)	=	[]		list-extension(p)	=	Lex
  prec(0)	=	0		status(0)	=	[]		list-extension(0)	=	Lex
  prec(if#)	=	0		status(if#)	=	[1, 2, 3]		list-extension(if#)	=	Lex

  and the following Max-polynomial interpretation

  [s(x1)]	=	x1 + 8856
  [le#(x1, x2)]	=	x2 + 1
  [minus(x1, x2)]	=	x2 + 1
  [false]	=	0
  [p#(x1)]	=	1
  [true]	=	0
  [p(x1)]	=	x1 + 0
  [0]	=	23676
  [if#(x1, x2, x3)]	=	max(x1 + 1, x2 + 1, x3 + 1, 0)

  together with the usable rules

  p(0)	→	0	(1)
  p(s(x))	→	x	(2)

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

  minus#(x,s(y))

  →

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

  (15)

  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

  le#(s(x),s(y))

  →

  le#(x,y)

  (11)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [le(x1, x2)]	=	0
  [s(x1)]	=	x1 + 1
  [le#(x1, x2)]	=	x1 + 0
  [minus(x1, x2)]	=	0
  [false]	=	0
  [p#(x1)]	=	0
  [true]	=	0
  [p(x1)]	=	0
  [0]	=	1
  [if(x1, x2, x3)]	=	0
  [minus#(x1, x2)]	=	0
  [if#(x1, x2, x3)]	=	0

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

  le#(s(x),s(y))

  →

  le#(x,y)

  (11)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.