Theorem 3dimlem1 37399

Description: Lemma for 3dim1 37408. (Contributed by NM, 25-Jul-2012.)

Hypotheses

Ref	Expression
3dim0.j	⊢ ∨ = (join‘𝐾)
3dim0.l	⊢ ≤ = (le‘𝐾)
3dim0.a	⊢ 𝐴 = (Atoms‘𝐾)

Assertion

Ref	Expression
3dimlem1	⊢ (((𝑄 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑄 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)) ∧ 𝑃 = 𝑄) → (𝑃 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑃 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆)))

Proof of Theorem 3dimlem1

Step	Hyp	Ref	Expression
1		neeq1 3005	. . 3 ⊢ (𝑃 = 𝑄 → (𝑃 ≠ 𝑅 ↔ 𝑄 ≠ 𝑅))
2		oveq1 7262	. . . . 5 ⊢ (𝑃 = 𝑄 → (𝑃 ∨ 𝑅) = (𝑄 ∨ 𝑅))
3	2	breq2d 5082	. . . 4 ⊢ (𝑃 = 𝑄 → (𝑆 ≤ (𝑃 ∨ 𝑅) ↔ 𝑆 ≤ (𝑄 ∨ 𝑅)))
4	3	notbid 317	. . 3 ⊢ (𝑃 = 𝑄 → (¬ 𝑆 ≤ (𝑃 ∨ 𝑅) ↔ ¬ 𝑆 ≤ (𝑄 ∨ 𝑅)))
5	2	oveq1d 7270	. . . . 5 ⊢ (𝑃 = 𝑄 → ((𝑃 ∨ 𝑅) ∨ 𝑆) = ((𝑄 ∨ 𝑅) ∨ 𝑆))
6	5	breq2d 5082	. . . 4 ⊢ (𝑃 = 𝑄 → (𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆) ↔ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)))
7	6	notbid 317	. . 3 ⊢ (𝑃 = 𝑄 → (¬ 𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆) ↔ ¬ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)))
8	1, 4, 7	3anbi123d 1434	. 2 ⊢ (𝑃 = 𝑄 → ((𝑃 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑃 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆)) ↔ (𝑄 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑄 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆))))
9	8	biimparc 479	1 ⊢ (((𝑄 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑄 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)) ∧ 𝑃 = 𝑄) → (𝑃 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑃 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   ∧ w3a 1085   = wceq 1539   ≠ wne 2942   class class class wbr 5070  ‘cfv 6418  (class class class)co 7255  lecple 16895  joincjn 17944  Atomscatm 37204

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ne 2943  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-uni 4837  df-br 5071  df-iota 6376  df-fv 6426  df-ov 7258

This theorem is referenced by:  3dim1  37408