Theorem orbile 825

Description: Disjunction of biconditionals.

Assertion

Ref	Expression
orbile	((a ≡ c) ∪ (b ≡ c)) ≤ (((a ∩ b) →2 c) ∩ (c →1 (a ∪ b)))

Proof of Theorem orbile

Step	Hyp	Ref	Expression
1		orbi 824	. 2 ((a ≡ c) ∪ (b ≡ c)) = (((a →2 c) ∪ (b →2 c)) ∩ ((c →1 a) ∪ (c →1 b)))
2		i2or 336	. . 3 ((a →2 c) ∪ (b →2 c)) ≤ ((a ∩ b) →2 c)
3		i1or 337	. . 3 ((c →1 a) ∪ (c →1 b)) ≤ (c →1 (a ∪ b))
4	2, 3	le2an 161	. 2 (((a →2 c) ∪ (b →2 c)) ∩ ((c →1 a) ∪ (c →1 b))) ≤ (((a ∩ b) →2 c) ∩ (c →1 (a ∪ b)))
5	1, 4	bltr 130	1 ((a ≡ c) ∪ (b ≡ c)) ≤ (((a ∩ b) →2 c) ∩ (c →1 (a ∪ b)))

Syntax hints:   ≤ wle 2   ≡ tb 5   ∪ wo 6   ∩ wa 7   →₁ wi1 13   →₂ wi2 14

This theorem is referenced by:  mlaconj4 826  mlaconj 827  mlaconjolem 867

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-r3 421

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  df-c1 124  df-c2 125

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