Theorem cdleme31se 41705

Description: Part of proof of Lemma D in [Crawley] p. 113. (Contributed by NM, 26-Feb-2013.)

Hypotheses

Ref	Expression
cdleme31se.e	⊢ 𝐸 = ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊)))
cdleme31se.y	⊢ 𝑌 = ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊)))

Assertion

Ref	Expression
cdleme31se	⊢ (𝑅 ∈ 𝐴 → ⦋𝑅 / 𝑠⦌𝐸 = 𝑌)

Distinct variable groups:   𝐴,𝑠   𝐷,𝑠   ∨ ,𝑠   ∧ ,𝑠   𝑃,𝑠   𝑄,𝑠   𝑅,𝑠   𝑊,𝑠   𝑇,𝑠

Allowed substitution hints:   𝐸(𝑠)   𝑌(𝑠)

Proof of Theorem cdleme31se

Step	Hyp	Ref	Expression
1		nfcvd 2927	. . 3 ⊢ (𝑅 ∈ 𝐴 → Ⅎ𝑠((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊))))
2		oveq1 7458	. . . . . 6 ⊢ (𝑠 = 𝑅 → (𝑠 ∨ 𝑇) = (𝑅 ∨ 𝑇))
3	2	oveq1d 7466	. . . . 5 ⊢ (𝑠 = 𝑅 → ((𝑠 ∨ 𝑇) ∧ 𝑊) = ((𝑅 ∨ 𝑇) ∧ 𝑊))
4	3	oveq2d 7467	. . . 4 ⊢ (𝑠 = 𝑅 → (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊)) = (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊)))
5	4	oveq2d 7467	. . 3 ⊢ (𝑠 = 𝑅 → ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊))) = ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊))))
6	1, 5	csbiegf 3884	. 2 ⊢ (𝑅 ∈ 𝐴 → ⦋𝑅 / 𝑠⦌((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊))) = ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊))))
7		cdleme31se.e	. . 3 ⊢ 𝐸 = ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊)))
8	7	csbeq2i 3859	. 2 ⊢ ⦋𝑅 / 𝑠⦌𝐸 = ⦋𝑅 / 𝑠⦌((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊)))
9		cdleme31se.y	. 2 ⊢ 𝑌 = ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊)))
10	6, 8, 9	3eqtr4g 2823	1 ⊢ (𝑅 ∈ 𝐴 → ⦋𝑅 / 𝑠⦌𝐸 = 𝑌)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∈ wcel 2147  ⦋csb 3851  (class class class)co 7451

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

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-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-br 5109  df-iota 6520  df-fv 6574  df-ov 7454

This theorem is used by:  cdleme31sde  41708  cdleme31sn1c  41711