Theorem axbnd 2601

Description: Axiom of Bundling (intuitionistic logic axiom ax-bnd). In classical logic, this and axi12 2600 are fairly straightforward consequences of axc9 2302. But in intuitionistic logic, it is not easy to add the extra A. x to axi12 2600 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 2318	. . . . . 6 |- F/ x -. A. z z = x
2		nfnae 2318	. . . . . 6 |- F/ x -. A. z z = y
3	1, 2	nfan 1828	. . . . 5 |- F/ x ( -. A. z z = x /\ -. A. z z = y )
4		nfnae 2318	. . . . . . 7 |- F/ z -. A. z z = x
5		nfnae 2318	. . . . . . 7 |- F/ z -. A. z z = y
6	4, 5	nfan 1828	. . . . . 6 |- F/ z ( -. A. z z = x /\ -. A. z z = y )
7		axc9 2302	. . . . . . 7 |- ( -. A. z z = x -> ( -. A. z z = y -> ( x = y -> A. z x = y ) ) )
8	7	imp 445	. . . . . 6 |- ( ( -. A. z z = x /\ -. A. z z = y ) -> ( x = y -> A. z x = y ) )
9	6, 8	alrimi 2082	. . . . 5 |- ( ( -. A. z z = x /\ -. A. z z = y ) -> A. z ( x = y -> A. z x = y ) )
10	3, 9	alrimi 2082	. . . 4 |- ( ( -. A. z z = x /\ -. A. z z = y ) -> A. x A. z ( x = y -> A. z x = y ) )
11	10	ex 450	. . 3 |- ( -. A. z z = x -> ( -. A. z z = y -> A. x A. z ( x = y -> A. z x = y ) ) )
12	11	orrd 393	. 2 |- ( -. A. z z = x -> ( A. z z = y \/ A. x A. z ( x = y -> A. z x = y ) ) )
13	12	orri 391	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 383   /\ wa 384   A.wal 1481

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710

This theorem is referenced by: (None)