let CX be Consistent Subset of CQC-WFF ; :: thesis: ( CX is with_examples implies ex CY being Consistent Subset of CQC-WFF st
( CX c= CY & CY is negation_faithful & CY is with_examples ) )
assume A1:
CX is with_examples
; :: thesis: ex CY being Consistent Subset of CQC-WFF st
( CX c= CY & CY is negation_faithful & CY is with_examples )
consider f being Function such that
A2:
( dom f = NAT & CQC-WFF = rng f )
by Lm1, Th19;
reconsider f = f as Function of NAT , CQC-WFF by A2, FUNCT_2:4;
defpred S1[ set , set , set ] means ex X being Subset of CQC-WFF st
( X = $2 \/ {(f . $1)} & ( X is Consistent implies $3 = X ) & ( not X is Consistent implies $3 = $2 ) );
A3:
for n being Element of NAT
for C being Subset of CQC-WFF ex D being Subset of CQC-WFF st S1[n,C,D]
consider h being Function of NAT , bool CQC-WFF such that
A4:
( h . 0 = CX & ( for n being Element of NAT holds S1[n,h . n,h . (n + 1)] ) )
from RECDEF_1:sch 2(A3);
set CY = union (rng h);
then A7:
CX c= union (rng h)
by TARSKI:def 3;
A8:
for n being Element of NAT holds h . n c= h . (n + 1)
A11:
for n, m being Element of NAT st m in dom h & n in dom h & n < m holds
h . n c= h . m
defpred S2[ Element of NAT ] means h . $1 is Consistent;
A18:
S2[ 0 ]
by A4;
A19:
for k being Element of NAT st S2[k] holds
S2[k + 1]
set CY = union (rng h);
for n being Element of NAT holds S2[n]
from NAT_1:sch 1(A18, A19);
then
for n, m being Element of NAT st m in dom h & n in dom h & n < m holds
( h . n is Consistent & h . n c= h . m )
by A11;
then reconsider CY = union (rng h) as Consistent Subset of CQC-WFF by HENMODEL:11;
A21:
CY is negation_faithful
A30:
CY is with_examples
proof
let x be
bound_QC-variable;
:: according to GOEDELCP:def 2 :: thesis: for p being Element of CQC-WFF ex y being bound_QC-variable st CY |- ('not' (Ex x,p)) 'or' (p . x,y)let p be
Element of
CQC-WFF ;
:: thesis: ex y being bound_QC-variable st CY |- ('not' (Ex x,p)) 'or' (p . x,y)
consider y being
bound_QC-variable such that A31:
CX |- ('not' (Ex x,p)) 'or' (p . x,y)
by A1, Def2;
take
y
;
:: thesis: CY |- ('not' (Ex x,p)) 'or' (p . x,y)
thus
CY |- ('not' (Ex x,p)) 'or' (p . x,y)
by A7, A31, Th32;
:: thesis: verum
end;
take
CY
; :: thesis: ( CX c= CY & CY is negation_faithful & CY is with_examples )
thus
( CX c= CY & CY is negation_faithful & CY is with_examples )
by A5, A21, A30, TARSKI:def 3; :: thesis: verum