let A, B be Element of LTLB_WFF ; for P being consistent complete PNPair
for Q being Element of compn P st A 'U' B in rng (P `2) holds
( B in rng (Q `2) & ( A in rng (Q `2) or A 'U' B in rng (Q `2) ) )
let P be consistent complete PNPair; for Q being Element of compn P st A 'U' B in rng (P `2) holds
( B in rng (Q `2) & ( A in rng (Q `2) or A 'U' B in rng (Q `2) ) )
let Q be Element of compn P; ( A 'U' B in rng (P `2) implies ( B in rng (Q `2) & ( A in rng (Q `2) or A 'U' B in rng (Q `2) ) ) )
set aub = A 'U' B;
set nb = 'not' B;
set na = 'not' A;
set au = A '&&' (A 'U' B);
('not' (untn (A,B))) => (('not' B) '&&' ('not' (A '&&' (A 'U' B)))) is ctaut
by LTLAXIO2:36;
then
('not' (untn (A,B))) => (('not' B) '&&' ('not' (A '&&' (A 'U' B)))) in LTL_axioms
by LTLAXIO1:def 17;
then A1:
{} LTLB_WFF |- ('not' (untn (A,B))) => (('not' B) '&&' ('not' (A '&&' (A 'U' B))))
by LTLAXIO1:42;
assume
A 'U' B in rng (P `2)
; ( B in rng (Q `2) & ( A in rng (Q `2) or A 'U' B in rng (Q `2) ) )
then A2:
untn (A,B) in rng (Q `2)
by Th21;
then A3:
untn (A,B) in rng Q
by XBOOLE_0:def 3;
then A4:
A in rng Q
by Th8;
{} LTLB_WFF |- (Q ^) => ('not' (untn (A,B)))
by LTLAXIO3:29, A2;
then A5:
{} LTLB_WFF |- (Q ^) => (('not' B) '&&' ('not' (A '&&' (A 'U' B))))
by A1, LTLAXIO1:47;
(('not' B) '&&' ('not' (A '&&' (A 'U' B)))) => ('not' B) is ctaut
by LTLAXIO2:27;
then
(('not' B) '&&' ('not' (A '&&' (A 'U' B)))) => ('not' B) in LTL_axioms
by LTLAXIO1:def 17;
then
{} LTLB_WFF |- (('not' B) '&&' ('not' (A '&&' (A 'U' B)))) => ('not' B)
by LTLAXIO1:42;
then A6:
{} LTLB_WFF |- (Q ^) => ('not' B)
by LTLAXIO1:47, A5;
(('not' B) '&&' ('not' (A '&&' (A 'U' B)))) => ('not' (A '&&' (A 'U' B))) is ctaut
by LTLAXIO2:28;
then
(('not' B) '&&' ('not' (A '&&' (A 'U' B)))) => ('not' (A '&&' (A 'U' B))) in LTL_axioms
by LTLAXIO1:def 17;
then
{} LTLB_WFF |- (('not' B) '&&' ('not' (A '&&' (A 'U' B)))) => ('not' (A '&&' (A 'U' B)))
by LTLAXIO1:42;
then A7:
{} LTLB_WFF |- (Q ^) => ('not' (A '&&' (A 'U' B)))
by LTLAXIO1:47, A5;
A8:
B in rng Q
by A3, Th8;
A9:
A 'U' B in rng Q
by A3, Th8;
assume A10:
( not B in rng (Q `2) or ( not A in rng (Q `2) & not A 'U' B in rng (Q `2) ) )
; contradiction