Theorem 3dimlem1 39497

Description: Lemma for 3dim1 39506. (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 2990	. . 3 ⊢ (𝑃 = 𝑄 → (𝑃 ≠ 𝑅 ↔ 𝑄 ≠ 𝑅))
2		oveq1 7348	. . . . 5 ⊢ (𝑃 = 𝑄 → (𝑃 ∨ 𝑅) = (𝑄 ∨ 𝑅))
3	2	breq2d 5098	. . . 4 ⊢ (𝑃 = 𝑄 → (𝑆 ≤ (𝑃 ∨ 𝑅) ↔ 𝑆 ≤ (𝑄 ∨ 𝑅)))
4	3	notbid 318	. . 3 ⊢ (𝑃 = 𝑄 → (¬ 𝑆 ≤ (𝑃 ∨ 𝑅) ↔ ¬ 𝑆 ≤ (𝑄 ∨ 𝑅)))
5	2	oveq1d 7356	. . . . 5 ⊢ (𝑃 = 𝑄 → ((𝑃 ∨ 𝑅) ∨ 𝑆) = ((𝑄 ∨ 𝑅) ∨ 𝑆))
6	5	breq2d 5098	. . . 4 ⊢ (𝑃 = 𝑄 → (𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆) ↔ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)))
7	6	notbid 318	. . 3 ⊢ (𝑃 = 𝑄 → (¬ 𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆) ↔ ¬ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)))
8	1, 4, 7	3anbi123d 1438	. 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 1086   = wceq 1541   ≠ wne 2928   class class class wbr 5086  ‘cfv 6476  (class class class)co 7341  lecple 17163  joincjn 18212  Atomscatm 39302

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-ss 3914  df-nul 4279  df-if 4471  df-sn 4572  df-pr 4574  df-op 4578  df-uni 4855  df-br 5087  df-iota 6432  df-fv 6484  df-ov 7344

This theorem is referenced by:  3dim1  39506