Theorem 3dimlem1 39441

Description: Lemma for 3dim1 39450. (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 3001	. . 3 ⊢ (𝑃 = 𝑄 → (𝑃 ≠ 𝑅 ↔ 𝑄 ≠ 𝑅))
2		oveq1 7438	. . . . 5 ⊢ (𝑃 = 𝑄 → (𝑃 ∨ 𝑅) = (𝑄 ∨ 𝑅))
3	2	breq2d 5160	. . . 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 7446	. . . . 5 ⊢ (𝑃 = 𝑄 → ((𝑃 ∨ 𝑅) ∨ 𝑆) = ((𝑄 ∨ 𝑅) ∨ 𝑆))
6	5	breq2d 5160	. . . 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 1435	. 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 1537   ≠ wne 2938   class class class wbr 5148  ‘cfv 6563  (class class class)co 7431  lecple 17305  joincjn 18369  Atomscatm 39245

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ne 2939  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-uni 4913  df-br 5149  df-iota 6516  df-fv 6571  df-ov 7434

This theorem is referenced by:  3dim1  39450