Certification Problem

Input (TPDB TRS_Equational/AProVE_AC_04/AC01)

The rewrite relation of the following equational TRS is considered.

Associative symbols: plus

Commutative symbols: plus

Property / Task

Answer / Result

Proof (by NaTT @ termCOMP 2023)

1 AC Dependency Pair Transformation

• The following set of (strict) dependency pairs is constructed for the TRS.

  plus#(x,s(y))

  →

  plus#(x,y)

  (8)

  Finiteness for these DPs in combination with the equational DPs is proven as follows.

  1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    plus#(x,y)	→	plus#(y,x)	(7)
    plus#(x,s(y))	→	plus#(x,y)	(8)
    plus#(x,plus(y,z))	→	plus#(x,y)	(6)
    plus#(x,plus(y,z))	→	plus#(plus(x,y),z)	(5)

    1.1.1 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	x1 + 1
    [plus#(x1, x2)]	=	x1 + x2 + 0
    [0]	=	21277
    [plus(x1, x2)]	=	x1 + x2 + 7720

    together with the usable rules

    plus(x,y)	→	plus(y,x)	(4)
    plus(x,0)	→	x	(1)
    plus(x,plus(y,z))	→	plus(plus(x,y),z)	(3)
    plus(x,s(y))	→	s(plus(x,y))	(2)

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

    plus#(x,plus(y,z))	→	plus#(x,y)	(6)
    plus#(x,s(y))	→	plus#(x,y)	(8)

    could be deleted.

    1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      plus#(x,y)	→	plus#(y,x)	(7)
      plus#(x,plus(y,z))	→	plus#(plus(x,y),z)	(5)

      1.1.1.1.1 AC Dependency Pair Problem is trivial

      There are no strict pairs and rules remaining, or there are no DPs remaining. Therefore, finiteness is trivially satisfied.

• The extended rules of the TRS

  plus(plus(x,s(y)),_1)	→	plus(s(plus(x,y)),_1)	(9)
  plus(plus(x,0),_1)	→	plus(x,_1)	(10)

  give rise to another dependency pair problem. Finiteness for these DPs in combination with the equational DPs is proven as follows.

  1.2 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    plus#(plus(x,0),_1)	→	plus#(x,_1)	(11)
    plus#(x,y)	→	plus#(y,x)	(7)
    plus#(x,plus(y,z))	→	plus#(plus(x,y),z)	(5)
    plus#(plus(x,s(y)),_1)	→	plus#(s(plus(x,y)),_1)	(12)
    plus#(x,plus(y,z))	→	plus#(x,y)	(6)

    1.2.1 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	2438
    [plus#(x1, x2)]	=	x1 + x2 + 0
    [0]	=	21277
    [plus(x1, x2)]	=	x1 + x2 + 1

    together with the usable rules

    plus(x,y)	→	plus(y,x)	(4)
    plus(x,0)	→	x	(1)
    plus(x,plus(y,z))	→	plus(plus(x,y),z)	(3)
    plus(x,s(y))	→	s(plus(x,y))	(2)

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

    plus#(x,plus(y,z))	→	plus#(x,y)	(6)
    plus#(plus(x,s(y)),_1)	→	plus#(s(plus(x,y)),_1)	(12)
    plus#(plus(x,0),_1)	→	plus#(x,_1)	(11)

    could be deleted.

    1.2.1.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      plus#(x,y)	→	plus#(y,x)	(7)
      plus#(x,plus(y,z))	→	plus#(plus(x,y),z)	(5)

      1.2.1.1.1 AC Dependency Pair Problem is trivial

      There are no strict pairs and rules remaining, or there are no DPs remaining. Therefore, finiteness is trivially satisfied.