Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/40708)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  1#(5(4(3(2(1(0(x1)))))))	→	1#(x1)	(7)
  1#(5(4(3(2(1(0(x1)))))))	→	1#(5(4(3(2(1(x1))))))	(8)
  1#(5(4(3(2(1(0(x1)))))))	→	1#(1(5(4(3(2(1(x1)))))))	(10)

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

  [5(x1)]	=	0 0 0 -3 0 0 -3 -3 0 · x1 + -∞ -∞ -∞
  [4(x1)]	=	0 0 0 -3 0 0 -3 -3 0 · x1 + -∞ -∞ -∞
  [3(x1)]	=	0 0 0 -3 -3 0 -3 -3 -3 · x1 + -∞ -∞ -∞
  [2(x1)]	=	0 0 0 -3 -3 0 -3 -3 0 · x1 + -∞ -∞ -∞
  [1(x1)]	=	0 0 0 -3 0 0 -3 -3 0 · x1 + -∞ -∞ -∞
  [0(x1)]	=	0 0 3 0 0 3 0 0 3 · x1 + -∞ -∞ -∞
  [1#(x1)]	=	15 16 17 15 16 17 15 16 17 · x1 + -∞ -∞ -∞

  together with the usable rules

  1(5(4(3(2(1(0(x1)))))))	→	5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(0(x1)))))))))))))))))))	(4)
  1(5(4(3(2(1(0(x1)))))))	→	5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1)))))))))))))))))))))))))	(5)
  1(5(4(3(2(1(0(x1)))))))	→	5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1)))))))))))))))))))))))))))))))	(6)

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

  1#(5(4(3(2(1(0(x1)))))))	→	1#(5(4(3(2(1(x1))))))	(8)
  1#(5(4(3(2(1(0(x1)))))))	→	1#(1(5(4(3(2(1(x1)))))))	(10)

  could be deleted.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    1#(5(4(3(2(1(0(x1)))))))

    →

    1#(x1)

    (7)

    1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

    Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1

    [5(x1)]	=	x1 + 1
    [4(x1)]	=	x1 + 1
    [3(x1)]	=	x1 + 1
    [2(x1)]	=	x1 + 1
    [1(x1)]	=	x1 + 1
    [0(x1)]	=	x1 + 1
    [1#(x1)]	=	x1 + 0

    having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

    1#(5(4(3(2(1(0(x1)))))))

    →

    1#(x1)

    (7)

    and no rules could be deleted.

    1.1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.