Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.35)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ 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

  ren#(x,y,lambda(z,t))	→	ren#(z,var(cons(x,cons(y,cons(lambda(z,t),nil)))),t)	(35)
  ren#(x,y,apply(t,s))	→	ren#(x,y,s)	(33)
  ren#(x,y,apply(t,s))	→	ren#(x,y,t)	(34)
  ren#(x,y,lambda(z,t))	→	ren#(x,y,ren(z,var(cons(x,cons(y,cons(lambda(z,t),nil)))),t))	(36)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the

  prec(ren#)	=	0		stat(ren#)	=	lex
  prec(ren)	=	0		stat(ren)	=	lex
  prec(if)	=	0		stat(if)	=	lex
  prec(lambda)	=	0		stat(lambda)	=	lex
  prec(apply)	=	0		stat(apply)	=	lex
  prec(var)	=	0		stat(var)	=	lex
  prec(cons)	=	0		stat(cons)	=	lex
  prec(eq)	=	0		stat(eq)	=	lex
  prec(nil)	=	0		stat(nil)	=	lex
  prec(false)	=	0		stat(false)	=	lex
  prec(and)	=	0		stat(and)	=	lex
  prec(true)	=	0		stat(true)	=	lex

  π(ren#)	=	3
  π(ren)	=	3
  π(if)	=	2
  π(lambda)	=	2
  π(apply)	=	[1,2]
  π(var)	=	[]
  π(cons)	=	[]
  π(eq)	=	[]
  π(nil)	=	[]
  π(false)	=	[]
  π(and)	=	2
  π(true)	=	[]

  together with the usable rules

  if(true,var(k),var(l'))	→	var(k)	(16)
  if(false,var(k),var(l'))	→	var(l')	(17)
  ren(var(l),var(k),var(l'))	→	if(eq(l,l'),var(k),var(l'))	(18)
  ren(x,y,apply(t,s))	→	apply(ren(x,y,t),ren(x,y,s))	(19)
  ren(x,y,lambda(z,t))	→	lambda(var(cons(x,cons(y,cons(lambda(z,t),nil)))),ren(x,y,ren(z,var(cons(x,cons(y,cons(lambda(z,t),nil)))),t)))	(20)

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

  ren#(x,y,apply(t,s))	→	ren#(x,y,s)	(33)
  ren#(x,y,apply(t,s))	→	ren#(x,y,t)	(34)

  could be deleted.

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the

  prec(ren#)	=	0		stat(ren#)	=	lex
  prec(ren)	=	0		stat(ren)	=	lex
  prec(if)	=	0		stat(if)	=	lex
  prec(lambda)	=	0		stat(lambda)	=	lex
  prec(apply)	=	0		stat(apply)	=	lex
  prec(var)	=	0		stat(var)	=	lex
  prec(cons)	=	0		stat(cons)	=	lex
  prec(eq)	=	0		stat(eq)	=	lex
  prec(nil)	=	0		stat(nil)	=	lex
  prec(false)	=	0		stat(false)	=	lex
  prec(and)	=	0		stat(and)	=	lex
  prec(true)	=	0		stat(true)	=	lex

  π(ren#)	=	3
  π(ren)	=	3
  π(if)	=	2
  π(lambda)	=	[2]
  π(apply)	=	1
  π(var)	=	[]
  π(cons)	=	2
  π(eq)	=	1
  π(nil)	=	[]
  π(false)	=	[]
  π(and)	=	[]
  π(true)	=	[]

  together with the usable rules

  if(true,var(k),var(l'))	→	var(k)	(16)
  if(false,var(k),var(l'))	→	var(l')	(17)
  ren(var(l),var(k),var(l'))	→	if(eq(l,l'),var(k),var(l'))	(18)
  ren(x,y,apply(t,s))	→	apply(ren(x,y,t),ren(x,y,s))	(19)
  ren(x,y,lambda(z,t))	→	lambda(var(cons(x,cons(y,cons(lambda(z,t),nil)))),ren(x,y,ren(z,var(cons(x,cons(y,cons(lambda(z,t),nil)))),t)))	(20)

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

  ren#(x,y,lambda(z,t))	→	ren#(z,var(cons(x,cons(y,cons(lambda(z,t),nil)))),t)	(35)
  ren#(x,y,lambda(z,t))	→	ren#(x,y,ren(z,var(cons(x,cons(y,cons(lambda(z,t),nil)))),t))	(36)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  eq#(cons(t,l),cons(t',l'))	→	eq#(l,l')	(21)
  eq#(cons(t,l),cons(t',l'))	→	eq#(t,t')	(22)
  eq#(var(l),var(l'))	→	eq#(l,l')	(24)
  eq#(apply(t,s),apply(t',s'))	→	eq#(s,s')	(25)
  eq#(apply(t,s),apply(t',s'))	→	eq#(t,t')	(26)
  eq#(lambda(x,t),lambda(x',t'))	→	eq#(t,t')	(28)
  eq#(lambda(x,t),lambda(x',t'))	→	eq#(x,x')	(29)

  1.1.2 Size-Change Termination

  Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

  eq#(cons(t,l),cons(t',l'))	→	eq#(l,l')	(21)
  2	>	2
  1	>	1
  eq#(cons(t,l),cons(t',l'))	→	eq#(t,t')	(22)
  2	>	2
  1	>	1
  eq#(var(l),var(l'))	→	eq#(l,l')	(24)
  2	>	2
  1	>	1
  eq#(apply(t,s),apply(t',s'))	→	eq#(s,s')	(25)
  2	>	2
  1	>	1
  eq#(apply(t,s),apply(t',s'))	→	eq#(t,t')	(26)
  2	>	2
  1	>	1
  eq#(lambda(x,t),lambda(x',t'))	→	eq#(t,t')	(28)
  2	>	2
  1	>	1
  eq#(lambda(x,t),lambda(x',t'))	→	eq#(x,x')	(29)
  2	>	2
  1	>	1

  As there is no critical graph in the transitive closure, there are no infinite chains.