Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/7)

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 1 component.

• The 1^st component contains the pair

  c#(c(c(a(x,y))))	→	c#(c(c(c(y))))	(10)
  c#(c(b(c(y),0)))	→	c#(c(a(y,0)))	(9)
  c#(c(a(a(y,0),x)))	→	c#(y)	(8)
  c#(c(c(a(x,y))))	→	c#(c(y))	(5)
  c#(c(c(a(x,y))))	→	c#(c(c(y)))	(7)
  c#(c(c(a(x,y))))	→	c#(y)	(6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1, x2)]	=	max(x1 + 0, x2 + 23160, 0)
  [b(x1, x2)]	=	max(x1 + 7720, x2 + 46320, 0)
  [c(x1)]	=	x1 + 15440
  [0]	=	54041
  [c#(x1)]	=	x1 + 0

  together with the usable rules

  c(c(c(a(x,y))))	→	b(c(c(c(c(y)))),x)	(1)
  c(c(a(a(y,0),x)))	→	c(y)	(3)
  c(c(b(c(y),0)))	→	a(0,c(c(a(y,0))))	(2)

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

  c#(c(c(a(x,y))))	→	c#(c(c(c(y))))	(10)
  c#(c(b(c(y),0)))	→	c#(c(a(y,0)))	(9)
  c#(c(a(a(y,0),x)))	→	c#(y)	(8)
  c#(c(c(a(x,y))))	→	c#(c(y))	(5)
  c#(c(c(a(x,y))))	→	c#(c(c(y)))	(7)
  c#(c(c(a(x,y))))	→	c#(y)	(6)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.