Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_CSR_05/Ex49_GM04)

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 3 components.

• The 1^st component contains the pair

  div#(s(X),s(Y))

  →

  div#(minus(X,Y),s(Y))

  (15)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [div#(x1, x2)]	=	x1 + 0
  [s(x1)]	=	x1 + 2
  [minus(x1, x2)]	=	x1 + 1
  [geq#(x1, x2)]	=	0
  [false]	=	0
  [div(x1, x2)]	=	0
  [geq(x1, x2)]	=	0
  [true]	=	0
  [0]	=	1
  [if(x1, x2, x3)]	=	0
  [minus#(x1, x2)]	=	0
  [if#(x1, x2, x3)]	=	0

  together with the usable rules

  minus(0,Y)	→	0	(1)
  minus(s(X),s(Y))	→	minus(X,Y)	(2)

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

  div#(s(X),s(Y))

  →

  div#(minus(X,Y),s(Y))

  (15)

  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

  geq#(s(X),s(Y))

  →

  geq#(X,Y)

  (11)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [div#(x1, x2)]	=	x1 + 0
  [s(x1)]	=	x1 + 2
  [minus(x1, x2)]	=	x1 + 1
  [geq#(x1, x2)]	=	x1 + x2 + 0
  [false]	=	0
  [div(x1, x2)]	=	0
  [geq(x1, x2)]	=	0
  [true]	=	0
  [0]	=	1
  [if(x1, x2, x3)]	=	0
  [minus#(x1, x2)]	=	0
  [if#(x1, x2, x3)]	=	0

  together with the usable rules

  minus(0,Y)	→	0	(1)
  minus(s(X),s(Y))	→	minus(X,Y)	(2)

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

  geq#(s(X),s(Y))

  →

  geq#(X,Y)

  (11)

  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

  minus#(s(X),s(Y))

  →

  minus#(X,Y)

  (14)

  1.1.3 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [div#(x1, x2)]	=	x1 + 0
  [s(x1)]	=	x1 + 2
  [minus(x1, x2)]	=	x1 + 1
  [geq#(x1, x2)]	=	0
  [false]	=	0
  [div(x1, x2)]	=	0
  [geq(x1, x2)]	=	0
  [true]	=	0
  [0]	=	1
  [if(x1, x2, x3)]	=	0
  [minus#(x1, x2)]	=	x1 + x2 + 0
  [if#(x1, x2, x3)]	=	0

  together with the usable rules

  minus(0,Y)	→	0	(1)
  minus(s(X),s(Y))	→	minus(X,Y)	(2)

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

  minus#(s(X),s(Y))

  →

  minus#(X,Y)

  (14)

  could be deleted.

  1.1.3.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.