Certification Problem

Input (TPDB TRS_Standard/Rubio_04/ma96)

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)	(37)
  ren#(X,Y,apply(T,S))	→	ren#(X,Y,S)	(35)
  ren#(X,Y,apply(T,S))	→	ren#(X,Y,T)	(36)
  ren#(X,Y,lambda(Z,T))	→	ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil)))),T))	(38)

  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(true)	=	0		stat(true)	=	lex
  prec(and)	=	0		stat(and)	=	lex
  prec(false)	=	0		stat(false)	=	lex

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

  together with the usable rules

  if(true,var(K),var(L))	→	var(K)	(18)
  if(false,var(K),var(L))	→	var(L)	(19)
  ren(var(L),var(K),var(Lp))	→	if(eq(L,Lp),var(K),var(Lp))	(20)
  ren(X,Y,apply(T,S))	→	apply(ren(X,Y,T),ren(X,Y,S))	(21)
  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)))	(22)

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

  ren#(X,Y,apply(T,S))	→	ren#(X,Y,S)	(35)
  ren#(X,Y,apply(T,S))	→	ren#(X,Y,T)	(36)

  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(true)	=	0		stat(true)	=	lex
  prec(and)	=	0		stat(and)	=	lex
  prec(false)	=	0		stat(false)	=	lex

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

  together with the usable rules

  if(true,var(K),var(L))	→	var(K)	(18)
  if(false,var(K),var(L))	→	var(L)	(19)
  ren(var(L),var(K),var(Lp))	→	if(eq(L,Lp),var(K),var(Lp))	(20)
  ren(X,Y,apply(T,S))	→	apply(ren(X,Y,T),ren(X,Y,S))	(21)
  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)))	(22)

  (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)	(37)
  ren#(X,Y,lambda(Z,T))	→	ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil)))),T))	(38)

  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(Tp,Lp))	→	eq#(L,Lp)	(23)
  eq#(cons(T,L),cons(Tp,Lp))	→	eq#(T,Tp)	(24)
  eq#(var(L),var(Lp))	→	eq#(L,Lp)	(26)
  eq#(apply(T,S),apply(Tp,Sp))	→	eq#(S,Sp)	(27)
  eq#(apply(T,S),apply(Tp,Sp))	→	eq#(T,Tp)	(28)
  eq#(lambda(X,T),lambda(Xp,Tp))	→	eq#(X,Xp)	(30)
  eq#(lambda(X,T),lambda(Xp,Tp))	→	eq#(T,Tp)	(31)

  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(Tp,Lp))	→	eq#(L,Lp)	(23)
  2	>	2
  1	>	1
  eq#(cons(T,L),cons(Tp,Lp))	→	eq#(T,Tp)	(24)
  2	>	2
  1	>	1
  eq#(var(L),var(Lp))	→	eq#(L,Lp)	(26)
  2	>	2
  1	>	1
  eq#(apply(T,S),apply(Tp,Sp))	→	eq#(S,Sp)	(27)
  2	>	2
  1	>	1
  eq#(apply(T,S),apply(Tp,Sp))	→	eq#(T,Tp)	(28)
  2	>	2
  1	>	1
  eq#(lambda(X,T),lambda(Xp,Tp))	→	eq#(X,Xp)	(30)
  2	>	2
  1	>	1
  eq#(lambda(X,T),lambda(Xp,Tp))	→	eq#(T,Tp)	(31)
  2	>	2
  1	>	1

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