Theorem 3dimlem1 40265

Description: Lemma for 3dim1 40274. (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 3022	. . 3 ⊢ (𝑃 = 𝑄 → (𝑃 ≠ 𝑅 ↔ 𝑄 ≠ 𝑅))
2		oveq1 7423	. . . . 5 ⊢ (𝑃 = 𝑄 → (𝑃 ∨ 𝑅) = (𝑄 ∨ 𝑅))
3	2	breq2d 5123	. . . 4 ⊢ (𝑃 = 𝑄 → (𝑆 ≤ (𝑃 ∨ 𝑅) ↔ 𝑆 ≤ (𝑄 ∨ 𝑅)))
4	3	notbid 321	. . 3 ⊢ (𝑃 = 𝑄 → (¬ 𝑆 ≤ (𝑃 ∨ 𝑅) ↔ ¬ 𝑆 ≤ (𝑄 ∨ 𝑅)))
5	2	oveq1d 7431	. . . . 5 ⊢ (𝑃 = 𝑄 → ((𝑃 ∨ 𝑅) ∨ 𝑆) = ((𝑄 ∨ 𝑅) ∨ 𝑆))
6	5	breq2d 5123	. . . 4 ⊢ (𝑃 = 𝑄 → (𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆) ↔ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)))
7	6	notbid 321	. . 3 ⊢ (𝑃 = 𝑄 → (¬ 𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆) ↔ ¬ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)))
8	1, 4, 7	3anbi123d 1464	. 2 ⊢ (𝑃 = 𝑄 → ((𝑃 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑃 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆)) ↔ (𝑄 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑄 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆))))
9	8	biimparc 485	1 ⊢ (((𝑄 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑄 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑄 ∨ 𝑅) ∨ 𝑆)) ∧ 𝑃 = 𝑄) → (𝑃 ≠ 𝑅 ∧ ¬ 𝑆 ≤ (𝑃 ∨ 𝑅) ∧ ¬ 𝑇 ≤ ((𝑃 ∨ 𝑅) ∨ 𝑆)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∧ w3a 1103   = wceq 1570   ≠ wne 2960   class class class wbr 5111  ‘cfv 6540  (class class class)co 7416  lecple 17335  joincjn 18385  Atomscatm 40070

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-uni 4875  df-br 5112  df-iota 6496  df-fv 6548  df-ov 7419

This theorem is used by:  3dim1  40274