Theorem cdleme31se 41258

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 2923	. . 3 ⊢ (𝑅 ∈ 𝐴 → Ⅎ𝑠((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊))))
2		oveq1 7421	. . . . . 6 ⊢ (𝑠 = 𝑅 → (𝑠 ∨ 𝑇) = (𝑅 ∨ 𝑇))
3	2	oveq1d 7429	. . . . 5 ⊢ (𝑠 = 𝑅 → ((𝑠 ∨ 𝑇) ∧ 𝑊) = ((𝑅 ∨ 𝑇) ∧ 𝑊))
4	3	oveq2d 7430	. . . 4 ⊢ (𝑠 = 𝑅 → (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊)) = (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊)))
5	4	oveq2d 7430	. . 3 ⊢ (𝑠 = 𝑅 → ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑠 ∨ 𝑇) ∧ 𝑊))) = ((𝑃 ∨ 𝑄) ∧ (𝐷 ∨ ((𝑅 ∨ 𝑇) ∧ 𝑊))))
6	1, 5	csbiegf 3880	. 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 3855	. 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 2820	1 ⊢ (𝑅 ∈ 𝐴 → ⦋𝑅 / 𝑠⦌𝐸 = 𝑌)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∈ wcel 2145  ⦋csb 3847  (class class class)co 7414

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 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732

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 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-rab 3413  df-v 3452  df-sbc 3740  df-csb 3848  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-iota 6489  df-fv 6541  df-ov 7417

This theorem is used by:  cdleme31sde  41261  cdleme31sn1c  41264