Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-22)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by NaTT @ 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

  c#(f(c(a,y,a)),x,z)	→	f#(b(b(z,z),f(b(y,b(x,a)))))	(15)
  b#(a,b(c(z,x,y),a))	→	c#(y,z,a)	(8)
  c#(f(c(a,y,a)),x,z)	→	b#(y,b(x,a))	(14)
  f#(c(a,b(b(z,a),y),x))	→	f#(c(x,b(z,x),y))	(13)
  f#(c(a,b(b(z,a),y),x))	→	b#(z,x)	(12)
  b#(a,b(c(z,x,y),a))	→	b#(z,c(y,z,a))	(7)
  c#(f(c(a,y,a)),x,z)	→	b#(z,z)	(5)
  c#(f(c(a,y,a)),x,z)	→	b#(b(z,z),f(b(y,b(x,a))))	(11)
  b#(a,b(c(z,x,y),a))	→	b#(b(z,c(y,z,a)),x)	(10)
  c#(f(c(a,y,a)),x,z)	→	f#(b(y,b(x,a)))	(9)
  f#(c(a,b(b(z,a),y),x))	→	c#(x,b(z,x),y)	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a]	=	1
  [b(x1, x2)]	=	21239
  [c(x1, x2, x3)]	=	x2 + 2
  [f(x1)]	=	1
  [f#(x1)]	=	x1 + 0
  [c#(x1, x2, x3)]	=	21240
  [b#(x1, x2)]	=	21240

  together with the usable rules

  b(a,b(c(z,x,y),a))	→	b(b(z,c(y,z,a)),x)	(1)
  c(f(c(a,y,a)),x,z)	→	f(b(b(z,z),f(b(y,b(x,a)))))	(3)
  f(c(a,b(b(z,a),y),x))	→	f(c(x,b(z,x),y))	(2)

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

  c#(f(c(a,y,a)),x,z)	→	f#(b(b(z,z),f(b(y,b(x,a)))))	(15)
  f#(c(a,b(b(z,a),y),x))	→	b#(z,x)	(12)
  c#(f(c(a,y,a)),x,z)	→	f#(b(y,b(x,a)))	(9)
  f#(c(a,b(b(z,a),y),x))	→	c#(x,b(z,x),y)	(4)

  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

    b#(a,b(c(z,x,y),a))	→	c#(y,z,a)	(8)
    b#(a,b(c(z,x,y),a))	→	b#(z,c(y,z,a))	(7)
    b#(a,b(c(z,x,y),a))	→	b#(b(z,c(y,z,a)),x)	(10)
    c#(f(c(a,y,a)),x,z)	→	b#(y,b(x,a))	(14)
    c#(f(c(a,y,a)),x,z)	→	b#(z,z)	(5)
    c#(f(c(a,y,a)),x,z)	→	b#(b(z,z),f(b(y,b(x,a))))	(11)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a]	=	0
    [b(x1, x2)]	=	max(x1 + 8862, x2 + 8861, 0)
    [c(x1, x2, x3)]	=	max(x1 + 8859, x2 + 8858, x3 + 8860, 0)
    [f(x1)]	=	8858
    [f#(x1)]	=	0
    [c#(x1, x2, x3)]	=	max(x1 + 2443, x2 + 11301, x3 + 11301, 0)
    [b#(x1, x2)]	=	max(x2 + 2438, 0)

    together with the usable rules

    b(a,b(c(z,x,y),a))	→	b(b(z,c(y,z,a)),x)	(1)
    c(f(c(a,y,a)),x,z)	→	f(b(b(z,z),f(b(y,b(x,a)))))	(3)
    f(c(a,b(b(z,a),y),x))	→	f(c(x,b(z,x),y))	(2)

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

    b#(a,b(c(z,x,y),a))	→	c#(y,z,a)	(8)
    b#(a,b(c(z,x,y),a))	→	b#(z,c(y,z,a))	(7)
    b#(a,b(c(z,x,y),a))	→	b#(b(z,c(y,z,a)),x)	(10)
    c#(f(c(a,y,a)),x,z)	→	b#(y,b(x,a))	(14)
    c#(f(c(a,y,a)),x,z)	→	b#(z,z)	(5)
    c#(f(c(a,y,a)),x,z)	→	b#(b(z,z),f(b(y,b(x,a))))	(11)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.

  • The 2^nd component contains the pair

    f#(c(a,b(b(z,a),y),x))

    →

    f#(c(x,b(z,x),y))

    (13)

    1.1.1.1.2 Reduction Pair Processor with Usable Rules

    Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

    [a]	=	0 1
    [b(x1, x2)]	=	1 1 0 0 · x1 + 1 0 0 0 · x2 + 22833 0
    [c(x1, x2, x3)]	=	1 1 0 0 · x1 + 0 0 1 0 · x2 + 0 1 1 0 · x3 + 2447 23368
    [f(x1)]	=	50255 1
    [f#(x1)]	=	0 1 0 0 · x1 + 0 0
    [c#(x1, x2, x3)]	=	0 0
    [b#(x1, x2)]	=	0 0

    together with the usable rules

    b(a,b(c(z,x,y),a))	→	b(b(z,c(y,z,a)),x)	(1)
    c(f(c(a,y,a)),x,z)	→	f(b(b(z,z),f(b(y,b(x,a)))))	(3)
    f(c(a,b(b(z,a),y),x))	→	f(c(x,b(z,x),y))	(2)

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

    f#(c(a,b(b(z,a),y),x))

    →

    f#(c(x,b(z,x),y))

    (13)

    could be deleted.

    1.1.1.1.2.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.