Theorem mlaconj2 846

Description: For 5GO proof of Mladen's conjecture. Hypothesis is 5GO law consequence. (Contributed by NM, 6-Jul-2000.)

Hypothesis

Ref	Expression
mlaconj2.1	((((a →1 (a ∩ b)) ∩ ((a ∩ b) →1 ((a ∩ b) ∪ c))) ∩ ((((a ∩ b) ∪ c) →1 c) ∩ (c →1 (a ∪ b)))) ∩ ((a ∪ b) →1 a)) ≤ (a ≡ c)

Assertion

Ref	Expression
mlaconj2	((a ≡ b) ∩ ((a ≡ c) ∪ (b ≡ c))) ≤ (a ≡ c)

Proof of Theorem mlaconj2

Step	Hyp	Ref	Expression
1		mlaconj 845	. 2 ((a ≡ b) ∩ ((a ≡ c) ∪ (b ≡ c))) ≤ ((((a →1 (a ∩ b)) ∩ ((a ∩ b) →1 ((a ∩ b) ∪ c))) ∩ ((((a ∩ b) ∪ c) →1 c) ∩ (c →1 (a ∪ b)))) ∩ ((a ∪ b) →1 a))
2		mlaconj2.1	. 2 ((((a →1 (a ∩ b)) ∩ ((a ∩ b) →1 ((a ∩ b) ∪ c))) ∩ ((((a ∩ b) ∪ c) →1 c) ∩ (c →1 (a ∪ b)))) ∩ ((a ∪ b) →1 a)) ≤ (a ≡ c)
3	1, 2	letr 137	1 ((a ≡ b) ∩ ((a ≡ c) ∪ (b ≡ c))) ≤ (a ≡ c)

This proof depends on syntax axioms:   ≤ wle 2   ≡ tb 5   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This proof depends on 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-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)