Theorem 3vth4 807

Description: A 3-variable theorem. (Contributed by NM, 18-Oct-1998.)

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 805	. . . 4 ((a →2 b) ∩ (b ∪ c)⊥ ) = ((a →2 c) ∩ (b ∪ c)⊥ )
2		ax-a1 30	. . . . 5 (a →2 b) = (a →2 b)⊥ ⊥
3	2	ran 78	. . . 4 ((a →2 b) ∩ (b ∪ c)⊥ ) = ((a →2 b)⊥ ⊥ ∩ (b ∪ c)⊥ )
4		ax-a1 30	. . . . 5 (a →2 c) = (a →2 c)⊥ ⊥
5	4	ran 78	. . . 4 ((a →2 c) ∩ (b ∪ c)⊥ ) = ((a →2 c)⊥ ⊥ ∩ (b ∪ c)⊥ )
6	1, 3, 5	3tr2 64	. . 3 ((a →2 b)⊥ ⊥ ∩ (b ∪ c)⊥ ) = ((a →2 c)⊥ ⊥ ∩ (b ∪ c)⊥ )
7	6	lor 70	. 2 ((b ∪ c) ∪ ((a →2 b)⊥ ⊥ ∩ (b ∪ c)⊥ )) = ((b ∪ c) ∪ ((a →2 c)⊥ ⊥ ∩ (b ∪ c)⊥ ))
8		df-i2 45	. 2 ((a →2 b)⊥ →2 (b ∪ c)) = ((b ∪ c) ∪ ((a →2 b)⊥ ⊥ ∩ (b ∪ c)⊥ ))
9		df-i2 45	. 2 ((a →2 c)⊥ →2 (b ∪ c)) = ((b ∪ c) ∪ ((a →2 c)⊥ ⊥ ∩ (b ∪ c)⊥ ))
10	7, 8, 9	3tr1 63	1 ((a →2 b)⊥ →2 (b ∪ c)) = ((a →2 c)⊥ →2 (b ∪ c))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 13

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i2 45  df-le1 130  df-le2 131

This theorem is used by:  3vth6  809