Theorem wdka4o 1116

Description: Show WDOL analog of WOM law. (Contributed by NM, 5-Mar-2006.)

Hypothesis

Ref	Expression
wdid0id5.1	(a ≡0 b) = 1

Assertion

Ref	Expression
wdka4o	((a ∪ c) ≡0 (b ∪ c)) = 1

Proof of Theorem wdka4o

Step	Hyp	Ref	Expression
1		wdid0id5.1	. . . 4 (a ≡0 b) = 1
2	1	wdid0id5 1111	. . 3 (a ≡ b) = 1
3	2	wr5 431	. 2 ((a ∪ c) ≡ (b ∪ c)) = 1
4	3	id5id0 352	1 ((a ∪ c) ≡0 (b ∪ c)) = 1

Syntax hints:   = wb 1   ∪ wo 6  1wt 8   ≡₀ wid0 17

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-wom 361  ax-wdol 1104

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45  df-id0 49  df-le 129  df-le1 130  df-le2 131  df-cmtr 134

This theorem is referenced by: (None)