Theorem xor2 1470

Description: Two ways to express "exclusive or." (Contributed by Mario Carneiro, 4-Sep-2016.)

Assertion

Ref	Expression
xor2	|- ( ( ph \/_ ps ) <-> ( ( ph \/ ps ) /\ -. ( ph /\ ps ) ) )

Proof of Theorem xor2

Step	Hyp	Ref	Expression
1		df-xor 1465	. 2 |- ( ( ph \/_ ps ) <-> -. ( ph <-> ps ) )
2		nbi2 936	. 2 |- ( -. ( ph <-> ps ) <-> ( ( ph \/ ps ) /\ -. ( ph /\ ps ) ) )
3	1, 2	bitri 264	1 |- ( ( ph \/_ ps ) <-> ( ( ph \/ ps ) /\ -. ( ph /\ ps ) ) )

Syntax hints:   -. wn 3   <-> wb 196   \/ wo 383   /\ wa 384   \/_ wxo 1464

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-xor 1465

This theorem is referenced by:  xoror  1471  xornan  1472  cador  1547  saddisjlem  15186  ifpdfxor  37832  dfxor4  38058  nanorxor  38504