Theorem cdleme31se 40384

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 2906	. . 3 ⊢ (𝑅 ∈ 𝐴 → Ⅎ𝑠((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊))))
2		oveq1 7438	. . . . . 6 ⊢ (𝑠 = 𝑅 → (𝑠 ∨ 𝑇) = (𝑅 ∨ 𝑇))
3	2	oveq1d 7446	. . . . 5 ⊢ (𝑠 = 𝑅 → ((𝑠 ∨ 𝑇) ∧ 𝑊) = ((𝑅 ∨ 𝑇) ∧ 𝑊))
4	3	oveq2d 7447	. . . 4 ⊢ (𝑠 = 𝑅 → (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊)) = (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊)))
5	4	oveq2d 7447	. . 3 ⊢ (𝑠 = 𝑅 → ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊))) = ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊))))
6	1, 5	csbiegf 3932	. 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 3907	. 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 2802	1 ⊢ (𝑅 ∈ 𝐴 → ⦋𝑅 / 𝑠⦌𝐸 = 𝑌)

Syntax hints:   → wi 4   = wceq 1540   ∈ wcel 2108  ⦋csb 3899  (class class class)co 7431

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-rab 3437  df-v 3482  df-sbc 3789  df-csb 3900  df-dif 3954  df-un 3956  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-uni 4908  df-br 5144  df-iota 6514  df-fv 6569  df-ov 7434

This theorem is referenced by:  cdleme31sde  40387  cdleme31sn1c  40390