Certification Problem

Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex14_Luc06_GM)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  a__g#(a,X)	→	a__f#(b,X)	(16)
  a__f#(X,X)	→	a__h#(a__a)	(17)
  a__h#(X)	→	a__g#(mark(X),X)	(14)
  a__h#(X)	→	mark#(X)	(15)
  mark#(h(X))	→	a__h#(mark(X))	(19)
  mark#(h(X))	→	mark#(X)	(20)
  mark#(g(X1,X2))	→	a__g#(mark(X1),X2)	(21)
  mark#(g(X1,X2))	→	mark#(X1)	(22)
  mark#(f(X1,X2))	→	a__f#(mark(X1),X2)	(24)
  mark#(f(X1,X2))	→	mark#(X1)	(25)

  1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [a__h(x1)]	=	2 + 2 · x1
  [a__g(x1, x2)]	=	2 + 1 · x1 + 1 · x2
  [mark(x1)]	=	1 · x1
  [a]	=	0
  [a__f(x1, x2)]	=	2 + 2 · x1 + 1 · x2
  [b]	=	0
  [a__a]	=	0
  [h(x1)]	=	2 + 2 · x1
  [g(x1, x2)]	=	2 + 1 · x1 + 1 · x2
  [f(x1, x2)]	=	2 + 2 · x1 + 1 · x2
  [a__g#(x1, x2)]	=	2 + 1 · x1 + 1 · x2
  [a__f#(x1, x2)]	=	2 + 2 · x1 + 1 · x2
  [a__h#(x1)]	=	2 + 2 · x1
  [mark#(x1)]	=	1 + 2 · x1

  the pairs

  a__h#(X)	→	mark#(X)	(15)
  mark#(h(X))	→	a__h#(mark(X))	(19)
  mark#(h(X))	→	mark#(X)	(20)
  mark#(g(X1,X2))	→	a__g#(mark(X1),X2)	(21)
  mark#(g(X1,X2))	→	mark#(X1)	(22)
  mark#(f(X1,X2))	→	a__f#(mark(X1),X2)	(24)
  mark#(f(X1,X2))	→	mark#(X1)	(25)

  and no rules could be deleted.

  1.1.1.1 Narrowing Processor

  We consider all narrowings of the pair below position 1 to get the following set of pairs

  a__f#(y0,y0)	→	a__h#(b)	(26)
  a__f#(y0,y0)	→	a__h#(a)	(27)

  1.1.1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  a__h#(b)	→	a__g#(mark(b),b)	(28)
  a__h#(a)	→	a__g#(mark(a),a)	(29)

  1.1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    a__f#(y0,y0)	→	a__h#(a)	(27)
    a__h#(a)	→	a__g#(mark(a),a)	(29)
    a__g#(a,X)	→	a__f#(b,X)	(16)

    1.1.1.1.1.1.1 Usable Rules Processor

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

    mark(a)	→	a__a	(7)
    a__a	→	b	(4)
    a__a	→	a	(12)

    1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

    a__a

    mark(h(x0))

    mark(g(x0,x1))

    mark(a)

    mark(f(x0,x1))

    mark(b)

    1.1.1.1.1.1.1.1.1 Narrowing Processor

    We consider all narrowings of the pair below position 1 to get the following set of pairs

    a__h#(a)

    →

    a__g#(a__a,a)

    (30)

    1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

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

    a__a	→	b	(4)
    a__a	→	a	(12)

    1.1.1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

    a__a

    1.1.1.1.1.1.1.1.1.1.1.1 Narrowing Processor

    We consider all narrowings of the pair below position 1 to get the following set of pairs

    a__h#(a)	→	a__g#(b,a)	(31)
    a__h#(a)	→	a__g#(a,a)	(32)

    1.1.1.1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      a__h#(a)	→	a__g#(a,a)	(32)
      a__g#(a,X)	→	a__f#(b,X)	(16)
      a__f#(y0,y0)	→	a__h#(a)	(27)

      1.1.1.1.1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

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

      There are no rules.

      1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

      There are no lhss.

      1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Instantiation Processor

      We instantiate the pair to the following set of pairs

      a__g#(a,a)

      →

      a__f#(b,a)

      (33)

      1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor

      The dependency pairs are split into 0 components.