Theorem w2or 354

Description: Join both sides with disjunction.

Hypotheses

Ref	Expression
w2or.1	(a ≡ b) = 1
w2or.2	(c ≡ d) = 1

Assertion

Ref	Expression
w2or	((a ∪ c) ≡ (b ∪ d)) = 1

Proof of Theorem w2or

Step	Hyp	Ref	Expression
1		w2or.2	. . 3 (c ≡ d) = 1
2	1	wlor 350	. 2 ((a ∪ c) ≡ (a ∪ d)) = 1
3		w2or.1	. . 3 (a ≡ b) = 1
4	3	wr5-2v 348	. 2 ((a ∪ d) ≡ (b ∪ d)) = 1
5	2, 4	wr2 353	1 ((a ∪ c) ≡ (b ∪ d)) = 1

Syntax hints:   = wb 1   ≡ tb 5   ∪ wo 6  1wt 9

This theorem is referenced by:  wcomlem 364  wdf-c1 365  wbctr 392  wcbtr 393  wcomcom5 402  wcomdr 403  wfh1 405  wcom2or 409  ska2 414

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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