Theorem w2an 355

Description: Join both sides with conjunction.

Hypotheses

Ref	Expression
w2an.1	(a ≡ b) = 1
w2an.2	(c ≡ d) = 1

Assertion

Ref	Expression
w2an	((a ∩ c) ≡ (b ∩ d)) = 1

Proof of Theorem w2an

Step	Hyp	Ref	Expression
1		w2an.2	. . 3 (c ≡ d) = 1
2	1	wlan 352	. 2 ((a ∩ c) ≡ (a ∩ d)) = 1
3		w2an.1	. . 3 (a ≡ b) = 1
4	3	wran 351	. 2 ((a ∩ d) ≡ (b ∩ d)) = 1
5	2, 4	wr2 353	1 ((a ∩ c) ≡ (b ∩ d)) = 1

Syntax hints:   = wb 1   ≡ tb 5   ∩ wa 7  1wt 9

This theorem is referenced by:  wcomd 400  wcom3ii 401  wcomcom5 402  wfh1 405  wfh3 407  wfh4 408

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-wom 343

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i1 43  df-i2 44  df-le1 122  df-le2 123

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