Certification Problem

Input (TPDB TRS_Innermost/AG01_innermost/#4.37a)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

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

  →

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

  (4)

  1.1.1.1 Usable Rules Processor

  We restrict the rewrite rules to the following usable rules of the DP problem.

  There are no rules.

  1.1.1.1.1 Innermost Lhss Removal Processor

  We restrict the innermost strategy to the following left hand sides.

  There are no lhss.

  1.1.1.1.1.1 Monotonic Reduction Pair Processor

  Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions

  prec(f#)	=	0		weight(f#)	=	1
  prec(s)	=	2		weight(s)	=	1
  prec(c)	=	1		weight(c)	=	0

  the pair

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

  →

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

  (4)

  and the rules

  f(c(s(x),y))	→	f(c(x,s(y)))	(1)
  g(c(x,s(y)))	→	g(c(s(x),y))	(2)

  could be deleted.

  1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  g#(c(x,s(y)))

  →

  g#(c(s(x),y))

  (5)

  1.1.1.2 Usable Rules Processor

  We restrict the rewrite rules to the following usable rules of the DP problem.

  There are no rules.

  1.1.1.2.1 Innermost Lhss Removal Processor

  We restrict the innermost strategy to the following left hand sides.

  There are no lhss.

  1.1.1.2.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [g#(x1)]	=	2 · x1
  [c(x1, x2)]	=	1 · x1 + 2 · x2
  [s(x1)]	=	2 + 1 · x1

  the pair

  g#(c(x,s(y)))

  →

  g#(c(s(x),y))

  (5)

  and the rules

  f(c(s(x),y))	→	f(c(x,s(y)))	(1)
  g(c(x,s(y)))	→	g(c(s(x),y))	(2)

  could be deleted.

  1.1.1.2.1.1.1 P is empty

  There are no pairs anymore.