Theorem oa4b 959

Description: Derivation of 4-OA law variant.

Hypothesis

Ref	Expression
oa4b.1	((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ (((a ∩ g) ∪ (c ∩ g)) ∪ (e ∩ g))

Assertion

Ref	Expression
oa4b	((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ g

Proof of Theorem oa4b

Step	Hyp	Ref	Expression
1		oa4b.1	. 2 ((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ (((a ∩ g) ∪ (c ∩ g)) ∪ (e ∩ g))
2		lear 161	. . . 4 (a ∩ g) ≤ g
3		lear 161	. . . 4 (c ∩ g) ≤ g
4	2, 3	lel2or 170	. . 3 ((a ∩ g) ∪ (c ∩ g)) ≤ g
5		lear 161	. . 3 (e ∩ g) ≤ g
6	4, 5	lel2or 170	. 2 (((a ∩ g) ∪ (c ∩ g)) ∪ (e ∩ g)) ≤ g
7	1, 6	letr 137	1 ((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ g

Syntax hints:   ≤ wle 2   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This theorem was proved from 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

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131

This theorem is referenced by: (None)