Certification Problem

Input (TPDB TRS_Equational/AProVE_AC_04/AC11)

The rewrite relation of the following equational TRS is considered.

Associative symbols: f

Commutative symbols: f

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.

  f#(g(h(a)),f(b,f(b,b)))	→	f#(h(a),a)	(10)
  f#(h(a),a)	→	f#(h(a),b)	(11)
  f#(g(f(h(a),a)),a)	→	f#(h(a),f(a,a))	(12)
  f#(g(h(a)),f(b,f(b,b)))	→	f#(g(f(h(a),a)),a)	(13)
  f#(h(a),g(a))	→	f#(g(h(a)),a)	(14)
  f#(g(f(h(a),a)),a)	→	f#(a,a)	(15)

  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

    f#(x,f(y,z))	→	f#(x,y)	(9)
    f#(g(f(h(a),a)),a)	→	f#(a,a)	(15)
    f#(h(a),g(a))	→	f#(g(h(a)),a)	(14)
    f#(x,f(y,z))	→	f#(f(x,y),z)	(8)
    f#(h(a),a)	→	f#(h(a),b)	(11)
    f#(g(h(a)),f(b,f(b,b)))	→	f#(h(a),a)	(10)
    f#(g(h(a)),f(b,f(b,b)))	→	f#(g(f(h(a),a)),a)	(13)
    f#(g(f(h(a),a)),a)	→	f#(h(a),f(a,a))	(12)
    f#(x,y)	→	f#(y,x)	(7)

    1.1.1 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [h(x1)]	=	11799
    [a]	=	1
    [b]	=	0
    [f(x1, x2)]	=	x1 + x2 + 1
    [f#(x1, x2)]	=	x1 + x2 + 0
    [g(x1)]	=	11802

    together with the usable rules

    f(h(a),a)	→	f(h(a),b)	(4)
    f(g(f(h(a),a)),a)	→	f(h(a),f(a,a))	(1)
    f(g(h(a)),f(b,f(b,b)))	→	f(g(f(h(a),a)),a)	(3)
    f(x,f(y,z))	→	f(f(x,y),z)	(6)
    f(x,y)	→	f(y,x)	(5)
    f(h(a),g(a))	→	f(g(h(a)),a)	(2)

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

    f#(g(f(h(a),a)),a)	→	f#(h(a),f(a,a))	(12)
    f#(g(h(a)),f(b,f(b,b)))	→	f#(g(f(h(a),a)),a)	(13)
    f#(g(h(a)),f(b,f(b,b)))	→	f#(h(a),a)	(10)
    f#(h(a),a)	→	f#(h(a),b)	(11)
    f#(h(a),g(a))	→	f#(g(h(a)),a)	(14)
    f#(g(f(h(a),a)),a)	→	f#(a,a)	(15)
    f#(x,f(y,z))	→	f#(x,y)	(9)

    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

      f#(x,f(y,z))	→	f#(f(x,y),z)	(8)
      f#(x,y)	→	f#(y,x)	(7)

      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

  f(f(h(a),g(a)),_1)	→	f(f(g(h(a)),a),_1)	(16)
  f(f(g(h(a)),f(b,f(b,b))),_1)	→	f(f(g(f(h(a),a)),a),_1)	(17)
  f(f(g(f(h(a),a)),a),_1)	→	f(f(h(a),f(a,a)),_1)	(18)
  f(f(h(a),a),_1)	→	f(f(h(a),b),_1)	(19)

  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

    f#(f(g(h(a)),f(b,f(b,b))),_1)	→	f#(f(g(f(h(a),a)),a),_1)	(20)
    f#(f(g(f(h(a),a)),a),_1)	→	f#(f(h(a),f(a,a)),_1)	(21)
    f#(x,f(y,z))	→	f#(f(x,y),z)	(8)
    f#(f(h(a),a),_1)	→	f#(f(h(a),b),_1)	(22)
    f#(x,y)	→	f#(y,x)	(7)
    f#(x,f(y,z))	→	f#(x,y)	(9)
    f#(f(h(a),g(a)),_1)	→	f#(f(g(h(a)),a),_1)	(23)

    1.2.1 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [h(x1)]	=	2
    [a]	=	1
    [b]	=	0
    [f(x1, x2)]	=	x1 + x2 + 8366
    [f#(x1, x2)]	=	x1 + x2 + 0
    [g(x1)]	=	8370

    together with the usable rules

    f(h(a),a)	→	f(h(a),b)	(4)
    f(g(f(h(a),a)),a)	→	f(h(a),f(a,a))	(1)
    f(g(h(a)),f(b,f(b,b)))	→	f(g(f(h(a),a)),a)	(3)
    f(x,f(y,z))	→	f(f(x,y),z)	(6)
    f(x,y)	→	f(y,x)	(5)
    f(h(a),g(a))	→	f(g(h(a)),a)	(2)

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

    f#(f(h(a),g(a)),_1)	→	f#(f(g(h(a)),a),_1)	(23)
    f#(x,f(y,z))	→	f#(x,y)	(9)
    f#(f(h(a),a),_1)	→	f#(f(h(a),b),_1)	(22)
    f#(f(g(f(h(a),a)),a),_1)	→	f#(f(h(a),f(a,a)),_1)	(21)
    f#(f(g(h(a)),f(b,f(b,b))),_1)	→	f#(f(g(f(h(a),a)),a),_1)	(20)

    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

      f#(x,f(y,z))	→	f#(f(x,y),z)	(8)
      f#(x,y)	→	f#(y,x)	(7)

      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.