Theorem 3vth4 789

Description: A 3-variable theorem.

Assertion

Ref	Expression
3vth4	((a →2 b)⊥ →2 (b ∪ c)) = ((a →2 c)⊥ →2 (b ∪ c))

Proof of Theorem 3vth4

Step	Hyp	Ref	Expression
1		3vth2 787	. . . 4 ((a →2 b) ∩ (b ∪ c)⊥ ) = ((a →2 c) ∩ (b ∪ c)⊥ )
2		ax-a1 29	. . . . 5 (a →2 b) = (a →2 b)⊥ ⊥
3	2	ran 71	. . . 4 ((a →2 b) ∩ (b ∪ c)⊥ ) = ((a →2 b)⊥ ⊥ ∩ (b ∪ c)⊥ )
4		ax-a1 29	. . . . 5 (a →2 c) = (a →2 c)⊥ ⊥
5	4	ran 71	. . . 4 ((a →2 c) ∩ (b ∪ c)⊥ ) = ((a →2 c)⊥ ⊥ ∩ (b ∪ c)⊥ )
6	1, 3, 5	3tr2 61	. . 3 ((a →2 b)⊥ ⊥ ∩ (b ∪ c)⊥ ) = ((a →2 c)⊥ ⊥ ∩ (b ∪ c)⊥ )
7	6	lor 66	. 2 ((b ∪ c) ∪ ((a →2 b)⊥ ⊥ ∩ (b ∪ c)⊥ )) = ((b ∪ c) ∪ ((a →2 c)⊥ ⊥ ∩ (b ∪ c)⊥ ))
8		df-i2 44	. 2 ((a →2 b)⊥ →2 (b ∪ c)) = ((b ∪ c) ∪ ((a →2 b)⊥ ⊥ ∩ (b ∪ c)⊥ ))
9		df-i2 44	. 2 ((a →2 c)⊥ →2 (b ∪ c)) = ((b ∪ c) ∪ ((a →2 c)⊥ ⊥ ∩ (b ∪ c)⊥ ))
10	7, 8, 9	3tr1 60	1 ((a →2 b)⊥ →2 (b ∪ c)) = ((a →2 c)⊥ →2 (b ∪ c))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 14

This theorem is referenced by:  3vth6 791

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

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

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