Theorem gomaex3h10 911

Description: Hypothesis for Godowski 6-var -> Mayet Example 3. (Contributed by NM, 29-Nov-1999.)

Hypotheses

Ref	Expression
gomaex3h10.10	q = ((e ∪ f) →1 (b ∪ c)⊥ )⊥
gomaex3h10.21	x = q
gomaex3h10.22	y = (e ∪ f)⊥

Assertion

Ref	Expression
gomaex3h10	x ≤ y⊥

Proof of Theorem gomaex3h10

Step	Hyp	Ref	Expression
1		lea 160	. . 3 ((e ∪ f) ∩ ((e ∪ f) ∩ (b ∪ c)⊥ )⊥ ) ≤ (e ∪ f)
2		gomaex3h10.10	. . . 4 q = ((e ∪ f) →1 (b ∪ c)⊥ )⊥
3		df-i1 44	. . . . . 6 ((e ∪ f) →1 (b ∪ c)⊥ ) = ((e ∪ f)⊥ ∪ ((e ∪ f) ∩ (b ∪ c)⊥ ))
4	3	ax-r4 37	. . . . 5 ((e ∪ f) →1 (b ∪ c)⊥ )⊥ = ((e ∪ f)⊥ ∪ ((e ∪ f) ∩ (b ∪ c)⊥ ))⊥
5		anor1 88	. . . . . 6 ((e ∪ f) ∩ ((e ∪ f) ∩ (b ∪ c)⊥ )⊥ ) = ((e ∪ f)⊥ ∪ ((e ∪ f) ∩ (b ∪ c)⊥ ))⊥
6	5	ax-r1 35	. . . . 5 ((e ∪ f)⊥ ∪ ((e ∪ f) ∩ (b ∪ c)⊥ ))⊥ = ((e ∪ f) ∩ ((e ∪ f) ∩ (b ∪ c)⊥ )⊥ )
7	4, 6	ax-r2 36	. . . 4 ((e ∪ f) →1 (b ∪ c)⊥ )⊥ = ((e ∪ f) ∩ ((e ∪ f) ∩ (b ∪ c)⊥ )⊥ )
8	2, 7	ax-r2 36	. . 3 q = ((e ∪ f) ∩ ((e ∪ f) ∩ (b ∪ c)⊥ )⊥ )
9		ax-a1 30	. . . 4 (e ∪ f) = (e ∪ f)⊥ ⊥
10	9	ax-r1 35	. . 3 (e ∪ f)⊥ ⊥ = (e ∪ f)
11	1, 8, 10	le3tr1 140	. 2 q ≤ (e ∪ f)⊥ ⊥
12		gomaex3h10.21	. 2 x = q
13		gomaex3h10.22	. . 3 y = (e ∪ f)⊥
14	13	ax-r4 37	. 2 y⊥ = (e ∪ f)⊥ ⊥
15	11, 12, 14	le3tr1 140	1 x ≤ y⊥

Syntax hints:   = wb 1   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-i1 44  df-le1 130  df-le2 131

This theorem is referenced by:  gomaex3lem5  918