Certification Problem

Input (TPDB TRS_Equational/Mixed_AC/YWHM14_4)

The rewrite relation of the following equational TRS is considered.

Associative symbols: ac1, ac2

Commutative symbols: ac1, ac2

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.

  ac2#(a,ac1(b,c))	→	ac1#(a,c)	(13)
  ac1#(a,ac2(b,c))	→	ac2#(a,c)	(14)
  ac2#(a,ac1(b,c))	→	ac2#(b,f(ac1(a,c)))	(15)
  ac1#(a,ac2(b,c))	→	ac1#(b,f(ac2(a,c)))	(16)

  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

    ac2#(x,ac2(y,z))	→	ac2#(ac2(x,y),z)	(12)
    ac1#(a,ac2(b,c))	→	ac1#(b,f(ac2(a,c)))	(16)
    ac2#(a,ac1(b,c))	→	ac2#(b,f(ac1(a,c)))	(15)
    ac2#(x,y)	→	ac2#(y,x)	(11)
    ac1#(a,ac2(b,c))	→	ac2#(a,c)	(14)
    ac1#(x,ac1(y,z))	→	ac1#(ac1(x,y),z)	(7)
    ac2#(x,ac2(y,z))	→	ac2#(x,y)	(10)
    ac1#(x,y)	→	ac1#(y,x)	(9)
    ac1#(x,ac1(y,z))	→	ac1#(x,y)	(8)
    ac2#(a,ac1(b,c))	→	ac1#(a,c)	(13)

    1.1.1 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a]	=	0
    [ac1#(x1, x2)]	=	2
    [b]	=	11292
    [c]	=	1
    [f(x1)]	=	3
    [ac1(x1, x2)]	=	x1 + x2 + 3
    [ac2(x1, x2)]	=	x1 + x2 + 1
    [ac2#(x1, x2)]	=	x1 + x2 + 0

    together with the usable rules

    ac2(x,ac2(y,z))	→	ac2(ac2(x,y),z)	(4)
    ac2(x,y)	→	ac2(y,x)	(3)
    ac2(a,ac1(b,c))	→	ac2(b,f(ac1(a,c)))	(2)

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

    ac2#(a,ac1(b,c))	→	ac1#(a,c)	(13)
    ac2#(x,ac2(y,z))	→	ac2#(x,y)	(10)
    ac1#(a,ac2(b,c))	→	ac2#(a,c)	(14)
    ac2#(a,ac1(b,c))	→	ac2#(b,f(ac1(a,c)))	(15)

    could be deleted.

    1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 2 components.

    • The 1^st component contains the pair

      ac1#(x,y)	→	ac1#(y,x)	(9)
      ac1#(a,ac2(b,c))	→	ac1#(b,f(ac2(a,c)))	(16)
      ac1#(x,ac1(y,z))	→	ac1#(x,y)	(8)
      ac1#(x,ac1(y,z))	→	ac1#(ac1(x,y),z)	(7)

      1.1.1.1.1 AC Reduction Pair Processor with Usable Rules

      Using the Max-polynomial interpretation

      [a]	=	0
      [ac1#(x1, x2)]	=	x1 + x2 + 2
      [b]	=	0
      [c]	=	1
      [f(x1)]	=	1
      [ac1(x1, x2)]	=	x1 + x2 + 1
      [ac2(x1, x2)]	=	x1 + x2 + 1
      [ac2#(x1, x2)]	=	x1 + x2 + 0

      together with the usable rules

      ac1(x,y)	→	ac1(y,x)	(6)
      ac1(a,ac2(b,c))	→	ac1(b,f(ac2(a,c)))	(1)
      ac1(x,ac1(y,z))	→	ac1(ac1(x,y),z)	(5)
      ac2(x,ac2(y,z))	→	ac2(ac2(x,y),z)	(4)
      ac2(x,y)	→	ac2(y,x)	(3)
      ac2(a,ac1(b,c))	→	ac2(b,f(ac1(a,c)))	(2)

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

      ac1#(x,ac1(y,z))	→	ac1#(x,y)	(8)
      ac1#(a,ac2(b,c))	→	ac1#(b,f(ac2(a,c)))	(16)

      could be deleted.

      1.1.1.1.1.1 Dependency Graph Processor

      The dependency pairs are split into 1 component.

      • The 1^st component contains the pair

        ac1#(x,y)	→	ac1#(y,x)	(9)
        ac1#(x,ac1(y,z))	→	ac1#(ac1(x,y),z)	(7)

        1.1.1.1.1.1.1 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 2^nd component contains the pair

      ac2#(x,ac2(y,z))	→	ac2#(ac2(x,y),z)	(12)
      ac2#(x,y)	→	ac2#(y,x)	(11)

      1.1.1.1.2 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

  ac2(ac2(a,ac1(b,c)),_1)	→	ac2(ac2(b,f(ac1(a,c))),_1)	(17)
  ac1(ac1(a,ac2(b,c)),_1)	→	ac1(ac1(b,f(ac2(a,c))),_1)	(18)

  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 2 components.

  • The 1^st component contains the pair

    ac2#(x,ac2(y,z))	→	ac2#(x,y)	(10)
    ac2#(x,y)	→	ac2#(y,x)	(11)
    ac2#(ac2(a,ac1(b,c)),_1)	→	ac2#(ac2(b,f(ac1(a,c))),_1)	(19)
    ac2#(x,ac2(y,z))	→	ac2#(ac2(x,y),z)	(12)

    1.2.1 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a]	=	0
    [ac1#(x1, x2)]	=	x1 + x2 + 2
    [b]	=	0
    [c]	=	2
    [f(x1)]	=	1
    [ac1(x1, x2)]	=	x1 + x2 + 0
    [ac2(x1, x2)]	=	x1 + x2 + 1
    [ac2#(x1, x2)]	=	x1 + x2 + 0

    together with the usable rules

    ac1(x,y)	→	ac1(y,x)	(6)
    ac1(a,ac2(b,c))	→	ac1(b,f(ac2(a,c)))	(1)
    ac1(x,ac1(y,z))	→	ac1(ac1(x,y),z)	(5)
    ac2(x,ac2(y,z))	→	ac2(ac2(x,y),z)	(4)
    ac2(x,y)	→	ac2(y,x)	(3)
    ac2(a,ac1(b,c))	→	ac2(b,f(ac1(a,c)))	(2)

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

    ac2#(ac2(a,ac1(b,c)),_1)	→	ac2#(ac2(b,f(ac1(a,c))),_1)	(19)
    ac2#(x,ac2(y,z))	→	ac2#(x,y)	(10)

    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

      ac2#(x,ac2(y,z))	→	ac2#(ac2(x,y),z)	(12)
      ac2#(x,y)	→	ac2#(y,x)	(11)

      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.
  • The 2^nd component contains the pair

    ac1#(x,y)	→	ac1#(y,x)	(9)
    ac1#(x,ac1(y,z))	→	ac1#(ac1(x,y),z)	(7)
    ac1#(x,ac1(y,z))	→	ac1#(x,y)	(8)
    ac1#(ac1(a,ac2(b,c)),_1)	→	ac1#(ac1(b,f(ac2(a,c))),_1)	(20)

    1.2.2 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a]	=	0
    [ac1#(x1, x2)]	=	x1 + x2 + 2
    [b]	=	1
    [c]	=	1
    [f(x1)]	=	1
    [ac1(x1, x2)]	=	x1 + x2 + 1
    [ac2(x1, x2)]	=	x1 + x2 + 1
    [ac2#(x1, x2)]	=	x1 + x2 + 0

    together with the usable rules

    ac1(x,y)	→	ac1(y,x)	(6)
    ac1(a,ac2(b,c))	→	ac1(b,f(ac2(a,c)))	(1)
    ac1(x,ac1(y,z))	→	ac1(ac1(x,y),z)	(5)
    ac2(x,ac2(y,z))	→	ac2(ac2(x,y),z)	(4)
    ac2(x,y)	→	ac2(y,x)	(3)
    ac2(a,ac1(b,c))	→	ac2(b,f(ac1(a,c)))	(2)

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

    ac1#(ac1(a,ac2(b,c)),_1)	→	ac1#(ac1(b,f(ac2(a,c))),_1)	(20)
    ac1#(x,ac1(y,z))	→	ac1#(x,y)	(8)

    could be deleted.

    1.2.2.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      ac1#(x,y)	→	ac1#(y,x)	(9)
      ac1#(x,ac1(y,z))	→	ac1#(ac1(x,y),z)	(7)

      1.2.2.1.1 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.