Theorem axbnd 2422

Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd).

In classical logic, this and axi12 2421 are fairly straightforward consequences of axc9 1994. But in intuitionistic logic, it is not easy to add the extra A. x to axi12 2421 and so we treat the two as separate axioms.

(Contributed by Jim Kingdon, 22-Mar-2018.)

Assertion

Ref	Expression
axbnd	|- ( A. z z = x \/ ( A. z z = y \/ A. x A. z ( x = y -> A. z x = y ) ) )

Proof of Theorem axbnd

Step	Hyp	Ref	Expression
1		nfnae 2005	. . . . . 6 |- F/ x -. A. z z = x
2		nfnae 2005	. . . . . 6 |- F/ x -. A. z z = y
3	1, 2	nfan 1861	. . . . 5 |- F/ x ( -. A. z z = x /\ -. A. z z = y )
4		nfnae 2005	. . . . . . 7 |- F/ z -. A. z z = x
5		nfnae 2005	. . . . . . 7 |- F/ z -. A. z z = y
6	4, 5	nfan 1861	. . . . . 6 |- F/ z ( -. A. z z = x /\ -. A. z z = y )
7		axc9 1994	. . . . . . 7 |- ( -. A. z z = x -> ( -. A. z z = y -> ( x = y -> A. z x = y ) ) )
8	7	imp 429	. . . . . 6 |- ( ( -. A. z z = x /\ -. A. z z = y ) -> ( x = y -> A. z x = y ) )
9	6, 8	alrimi 1811	. . . . 5 |- ( ( -. A. z z = x /\ -. A. z z = y ) -> A. z ( x = y -> A. z x = y ) )
10	3, 9	alrimi 1811	. . . 4 |- ( ( -. A. z z = x /\ -. A. z z = y ) -> A. x A. z ( x = y -> A. z x = y ) )
11	10	ex 434	. . 3 |- ( -. A. z z = x -> ( -. A. z z = y -> A. x A. z ( x = y -> A. z x = y ) ) )
12	11	orrd 378	. 2 |- ( -. A. z z = x -> ( A. z z = y \/ A. x A. z ( x = y -> A. z x = y ) ) )
13	12	orri 376	1 |- ( A. z z = x \/ ( A. z z = y \/ A. x A. z ( x = y -> A. z x = y ) ) )

Syntax hints:   -. wn 3   -> wi 4   \/ wo 368   /\ wa 369   A.wal 1367

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1591  ax-4 1602  ax-5 1670  ax-6 1708  ax-7 1728  ax-10 1775  ax-11 1780  ax-12 1792  ax-13 1943

This theorem depends on definitions:  df-bi 185  df-or 370  df-an 371  df-ex 1587  df-nf 1590

This theorem is referenced by: (None)