Certification Problem

Input (TPDB TRS_Standard/AG01/#3.54)

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

  f'#(s(x),y,y)

  →

  f'#(y,x,s(x))

  (6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [h(x1)]	=	0
  [s(x1)]	=	x1 + 1
  [f'#(x1, x2, x3)]	=	x1 + x2 + 0
  [f(x1)]	=	0
  [f#(x1)]	=	0
  [f'(x1, x2, x3)]	=	0
  [g(x1)]	=	0

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

  f'#(s(x),y,y)

  →

  f'#(y,x,s(x))

  (6)

  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#(g(x))	→	f#(x)	(5)
  f#(g(x))	→	f#(f(x))	(4)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [h(x1)]	=	1
  [s(x1)]	=	1
  [f'#(x1, x2, x3)]	=	0
  [f(x1)]	=	x1 + 0
  [f#(x1)]	=	x1 + 0
  [f'(x1, x2, x3)]	=	0
  [g(x1)]	=	x1 + 8856

  together with the usable rules

  f(g(x))	→	g(f(f(x)))	(1)
  f(h(x))	→	h(g(x))	(2)

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

  f#(g(x))	→	f#(x)	(5)
  f#(g(x))	→	f#(f(x))	(4)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.