Certification Problem

Input (TPDB TRS_Standard/AG01/#3.36)

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

  g#(s(x))	→	g#(x)	(13)
  g#(s(x))	→	f#(g(x))	(11)
  f#(s(x))	→	g#(f(x))	(8)
  f#(s(x))	→	f#(x)	(9)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 23092
  [minus(x1, x2)]	=	x1 + 0
  [f(x1)]	=	x1 + 25529
  [0]	=	1
  [f#(x1)]	=	x1 + 2438
  [g#(x1)]	=	x1 + 0
  [minus#(x1, x2)]	=	0
  [g(x1)]	=	x1 + 20653

  together with the usable rules

  f(s(x))	→	minus(s(x),g(f(x)))	(4)
  minus(x,0)	→	x	(1)
  f(0)	→	s(0)	(3)
  g(0)	→	0	(5)
  g(s(x))	→	minus(s(x),f(g(x)))	(6)
  minus(s(x),s(y))	→	minus(x,y)	(2)

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

  g#(s(x))	→	g#(x)	(13)
  g#(s(x))	→	f#(g(x))	(11)
  f#(s(x))	→	g#(f(x))	(8)
  f#(s(x))	→	f#(x)	(9)

  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

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

  →

  minus#(x,y)

  (12)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 1
  [minus(x1, x2)]	=	x1 + 0
  [f(x1)]	=	x1 + 1
  [0]	=	1
  [f#(x1)]	=	x1 + 2438
  [g#(x1)]	=	x1 + 0
  [minus#(x1, x2)]	=	x2 + 0
  [g(x1)]	=	x1 + 36466

  together with the usable rules

  f(s(x))	→	minus(s(x),g(f(x)))	(4)
  minus(x,0)	→	x	(1)
  f(0)	→	s(0)	(3)
  g(0)	→	0	(5)
  g(s(x))	→	minus(s(x),f(g(x)))	(6)
  minus(s(x),s(y))	→	minus(x,y)	(2)

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

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

  →

  minus#(x,y)

  (12)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.