Certification Problem

Input (TPDB TRS_Equational/Mixed_AC_and_C/AC29)

The rewrite relation of the following equational TRS is considered.

Associative symbols: app

Commutative symbols: max', app

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.

  6#	→	5#	(25)
  max2#(x,singl(y))	→	max'#(x,y)	(26)
  max2#(x,app(singl(y),Z))	→	max2#(max'(x,y),Z)	(27)
  3#	→	2#	(28)
  7#	→	6#	(29)
  max2#(x,app(singl(y),Z))	→	max'#(x,y)	(30)
  max#(app(singl(x),Y))	→	max2#(x,Y)	(31)
  4#	→	3#	(32)
  9#	→	8#	(33)
  2#	→	1#	(34)
  5#	→	4#	(35)
  8#	→	7#	(36)
  max'#(s(x),s(y))	→	max'#(x,y)	(37)

  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 3 components.

  • The 1^st component contains the pair

    app#(x,app(y,z))	→	app#(x,y)	(23)
    app#(x,app(y,z))	→	app#(app(x,y),z)	(22)
    app#(x,y)	→	app#(y,x)	(21)

    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

    max2#(x,app(singl(y),Z))

    →

    max2#(max'(x,y),Z)

    (27)

    1.1.2 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [7]	=	0
    [1]	=	0
    [4]	=	0
    [s(x1)]	=	x1 + 1
    [7#]	=	0
    [max2(x1, x2)]	=	0
    [max'(x1, x2)]	=	x1 + x2 + 1
    [5]	=	0
    [max'#(x1, x2)]	=	0
    [3]	=	0
    [6#]	=	0
    [8#]	=	0
    [2#]	=	0
    [9]	=	0
    [max2#(x1, x2)]	=	x1 + x2 + 0
    [8]	=	0
    [4#]	=	0
    [9#]	=	0
    [0]	=	10157
    [max(x1)]	=	0
    [max#(x1)]	=	0
    [3#]	=	0
    [app#(x1, x2)]	=	0
    [singl(x1)]	=	x1 + 2
    [5#]	=	0
    [2]	=	0
    [empty]	=	0
    [6]	=	0
    [1#]	=	0
    [app(x1, x2)]	=	x1 + x2 + 11797

    together with the usable rules

    max'(x,y)	→	max'(y,x)	(20)
    max'(0,x)	→	x	(10)
    max'(s(x),s(y))	→	s(max'(x,y))	(11)

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

    max2#(x,app(singl(y),Z))

    →

    max2#(max'(x,y),Z)

    (27)

    could be deleted.

    1.1.2.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.

  • The 3^rd component contains the pair

    max'#(s(x),s(y))	→	max'#(x,y)	(37)
    max'#(x,y)	→	max'#(y,x)	(24)

    1.1.3 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [7]	=	0
    [1]	=	0
    [4]	=	0
    [s(x1)]	=	x1 + 1
    [7#]	=	0
    [max2(x1, x2)]	=	0
    [max'(x1, x2)]	=	x1 + x2 + 1
    [5]	=	0
    [max'#(x1, x2)]	=	x1 + x2 + 0
    [3]	=	0
    [6#]	=	0
    [8#]	=	0
    [2#]	=	0
    [9]	=	0
    [max2#(x1, x2)]	=	x1 + 0
    [8]	=	0
    [4#]	=	0
    [9#]	=	0
    [0]	=	10157
    [max(x1)]	=	0
    [max#(x1)]	=	0
    [3#]	=	0
    [app#(x1, x2)]	=	0
    [singl(x1)]	=	2
    [5#]	=	0
    [2]	=	0
    [empty]	=	0
    [6]	=	0
    [1#]	=	0
    [app(x1, x2)]	=	11797

    together with the usable rules

    max'(x,y)	→	max'(y,x)	(20)
    max'(0,x)	→	x	(10)
    max'(s(x),s(y))	→	s(max'(x,y))	(11)

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

    max'#(s(x),s(y))

    →

    max'#(x,y)

    (37)

    could be deleted.

    1.1.3.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      max'#(x,y)

      →

      max'#(y,x)

      (24)

      1.1.3.1.1 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

  app(app(empty,X),_1)

  →

  app(X,_1)

  (38)

  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

    max'#(x,y)

    →

    max'#(y,x)

    (24)

    1.2.1 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

    app#(app(empty,X),_1)	→	app#(X,_1)	(39)
    app#(x,app(y,z))	→	app#(x,y)	(23)
    app#(x,y)	→	app#(y,x)	(21)
    app#(x,app(y,z))	→	app#(app(x,y),z)	(22)

    1.2.2 AC Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [7]	=	0
    [1]	=	0
    [4]	=	0
    [s(x1)]	=	x1 + 1
    [7#]	=	0
    [max2(x1, x2)]	=	0
    [max'(x1, x2)]	=	x1 + x2 + 1
    [5]	=	0
    [max'#(x1, x2)]	=	0
    [3]	=	0
    [6#]	=	0
    [8#]	=	0
    [2#]	=	0
    [9]	=	0
    [max2#(x1, x2)]	=	x1 + 0
    [8]	=	0
    [4#]	=	0
    [9#]	=	0
    [0]	=	10157
    [max(x1)]	=	0
    [max#(x1)]	=	0
    [3#]	=	0
    [app#(x1, x2)]	=	x1 + x2 + 0
    [singl(x1)]	=	2
    [5#]	=	0
    [2]	=	0
    [empty]	=	40651
    [6]	=	0
    [1#]	=	0
    [app(x1, x2)]	=	x1 + x2 + 11797

    together with the usable rules

    max'(x,y)	→	max'(y,x)	(20)
    app(x,app(y,z))	→	app(app(x,y),z)	(19)
    max'(0,x)	→	x	(10)
    app(x,y)	→	app(y,x)	(18)
    app(empty,X)	→	X	(12)
    max'(s(x),s(y))	→	s(max'(x,y))	(11)

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

    app#(x,app(y,z))	→	app#(x,y)	(23)
    app#(app(empty,X),_1)	→	app#(X,_1)	(39)

    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

      app#(x,app(y,z))	→	app#(app(x,y),z)	(22)
      app#(x,y)	→	app#(y,x)	(21)

      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.