Theorem hbequid 1501

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-8 1492 and ax-i12 1495 on top of (the FOL analogue of) modal logic KT. This shows that this can be proved without ax-i9 1518, even though Theorem equid 1689 cannot. A shorter proof using ax-i9 1518 is obtainable from equid 1689 and hbth 1451. (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 1496	. 2 |- ( A. y y = x \/ ( A. y y = x \/ A. y ( x = x -> A. y x = x ) ) )
2		ax-8 1492	. . . . . 6 |- ( y = x -> ( y = x -> x = x ) )
3	2	pm2.43i 49	. . . . 5 |- ( y = x -> x = x )
4	3	alimi 1443	. . . 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 1498	. . . 4 |- ( A. y ( x = x -> A. y x = x ) -> ( x = x -> A. y x = x ) )
7	5, 6	jaoi 706	. . 3 |- ( ( A. y y = x \/ A. y ( x = x -> A. y x = x ) ) -> ( x = x -> A. y x = x ) )
8	5, 7	jaoi 706	. 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 5	1 |- ( x = x -> A. y x = x )

Syntax hints:   -> wi 4   \/ wo 698   A.wal 1341   = wceq 1343

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-gen 1437  ax-8 1492  ax-i12 1495  ax-4 1498

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  equveli  1747