Certification Problem

Input (TPDB TRS_Standard/Rubio_04/bintrees)

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 2 components.

• The 1^st component contains the pair

  lessleaves#(cons(U,V),cons(W,Z))

  →

  lessleaves#(concat(U,V),concat(W,Z))

  (6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [concat#(x1, x2)]	=	0
  [false]	=	0
  [lessleaves#(x1, x2)]	=	x1 + x2 + 0
  [true]	=	0
  [concat(x1, x2)]	=	x1 + x2 + 1
  [cons(x1, x2)]	=	x1 + x2 + 2
  [lessleaves(x1, x2)]	=	0
  [leaf]	=	20163

  together with the usable rules

  concat(leaf,Y)	→	Y	(1)
  concat(cons(U,V),Y)	→	cons(U,concat(V,Y))	(2)

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

  lessleaves#(cons(U,V),cons(W,Z))

  →

  lessleaves#(concat(U,V),concat(W,Z))

  (6)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  concat#(cons(U,V),Y)

  →

  concat#(V,Y)

  (8)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [concat#(x1, x2)]	=	x1 + 0
  [false]	=	0
  [lessleaves#(x1, x2)]	=	x1 + x2 + 0
  [true]	=	0
  [concat(x1, x2)]	=	x1 + x2 + 1
  [cons(x1, x2)]	=	x1 + x2 + 2
  [lessleaves(x1, x2)]	=	0
  [leaf]	=	42736

  together with the usable rules

  concat(leaf,Y)	→	Y	(1)
  concat(cons(U,V),Y)	→	cons(U,concat(V,Y))	(2)

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

  concat#(cons(U,V),Y)

  →

  concat#(V,Y)

  (8)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.