Theorem hbequid 1406

Description: Bound-variable hypothesis builder for x = x. This theorem tells us that any variable, including x, is effectively not free in x = x, even though x is technically free according to the traditional definition of free variable. (The proof uses only ax-5 1336, ax-8 1395, ax-12 1402, and ax-gen 1338. This shows that this can be proved without ax-9 1424, even though the theorem equid 1589 cannot be. A shorter proof using ax-9 1424 is obtainable from equid 1589 and hbth 1352.) (Contributed by NM, 13-Jan-2011.) (Proof shortened by Wolf Lammen, 23-Mar-2014.)

Assertion

Ref	Expression
hbequid	|- ( x = x -> A. y x = x )

Proof of Theorem hbequid

Step	Hyp	Ref	Expression
1		ax12or 1403	. 2 |- ( A. y y = x \/ ( A. y y = x \/ A. y ( x = x -> A. y x = x ) ) )
2		ax-8 1395	. . . . . 6 |- ( y = x -> ( y = x -> x = x ) )
3	2	pm2.43i 43	. . . . 5 |- ( y = x -> x = x )
4	3	alimi 1344	. . . 4 |- ( A. y y = x -> A. y x = x )
5	4	a1d 22	. . 3 |- ( A. y y = x -> ( x = x -> A. y x = x ) )
6		ax-4 1400	. . . 4 |- ( A. y ( x = x -> A. y x = x ) -> ( x = x -> A. y x = x ) )
7	5, 6	jaoi 636	. . 3 |- ( ( A. y y = x \/ A. y ( x = x -> A. y x = x ) ) -> ( x = x -> A. y x = x ) )
8	5, 7	jaoi 636	. 2 |- ( ( A. y y = x \/ ( A. y y = x \/ A. y ( x = x -> A. y x = x ) ) ) -> ( x = x -> A. y x = x ) )
9	1, 8	ax-mp 7	1 |- ( x = x -> A. y x = x )

Syntax hints:   -> wi 4   \/ wo 629   A.wal 1241   = wceq 1243

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-gen 1338  ax-8 1395  ax-i12 1398  ax-4 1400

This theorem depends on definitions:  df-bi 110

This theorem is referenced by:  equveli  1642