Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.34)

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#(x,c(y))	→	g#(s(c(y)),y)	(12)
  g#(x,c(y))	→	g#(x,g(s(c(y)),y))	(10)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [1]	=	1
  [s(x1)]	=	1
  [false]	=	41063
  [c(x1)]	=	x1 + 41065
  [true]	=	41063
  [f(x1)]	=	41063
  [0]	=	1
  [if(x1, x2, x3)]	=	x1 + x2 + x3 + 0
  [f#(x1)]	=	0
  [g#(x1, x2)]	=	x2 + 0
  [if#(x1, x2, x3)]	=	0
  [g(x1, x2)]	=	x1 + x2 + 41063

  together with the usable rules

  if(true,x,y)	→	x	(4)
  f(0)	→	true	(1)
  f(s(x))	→	f(x)	(3)
  if(false,x,y)	→	y	(5)
  g(x,c(y))	→	g(x,g(s(c(y)),y))	(7)
  g(s(x),s(y))	→	if(f(x),s(x),s(y))	(6)
  f(1)	→	false	(2)

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

  g#(x,c(y))	→	g#(s(c(y)),y)	(12)
  g#(x,c(y))	→	g#(x,g(s(c(y)),y))	(10)

  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

  f#(s(x))

  →

  f#(x)

  (9)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [1]	=	0
  [s(x1)]	=	x1 + 1
  [false]	=	2
  [c(x1)]	=	x1 + 0
  [true]	=	1
  [f(x1)]	=	x1 + 2
  [0]	=	0
  [if(x1, x2, x3)]	=	x1 + x2 + 0
  [f#(x1)]	=	x1 + 0
  [g#(x1, x2)]	=	0
  [if#(x1, x2, x3)]	=	0
  [g(x1, x2)]	=	x1 + x2 + 0

  together with the usable rules

  f(0)	→	true	(1)
  f(s(x))	→	f(x)	(3)
  f(1)	→	false	(2)

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

  f#(s(x))

  →

  f#(x)

  (9)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.