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Theorem rabxfrd 5378
Description: Membership in a restricted class abstraction after substituting an expression 𝐴 (containing 𝑦) for 𝑥 in the formula defining the class abstraction. (Contributed by NM, 16-Jan-2012.)
Hypotheses
Ref Expression
rabxfrd.1 Ⅎ𝑦𝐵
rabxfrd.2 Ⅎ𝑦𝐶
rabxfrd.3 ((𝜑 ∧ 𝑦 ∈ 𝐷) → 𝐴 ∈ 𝐷)
rabxfrd.4 (𝑥 = 𝐴 → (𝜓 ↔ 𝜒))
rabxfrd.5 (𝑦 = 𝐵 → 𝐴 = 𝐶)
Assertion
Ref Expression
rabxfrd ((𝜑 ∧ 𝐵 ∈ 𝐷) → (𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}))
Distinct variable groups:   𝑥,𝐴   𝑥,𝑦,𝐷   𝜑,𝑦   𝜓,𝑦   𝜒,𝑥
Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)   𝜒(𝑦)   𝐴(𝑦)   𝐵(𝑥, 𝑦)   𝐶(𝑥, 𝑦)

Proof of Theorem rabxfrd
StepHypRef Expression
1 rabxfrd.3 . . . . . . . . . . 11 ((𝜑 ∧ 𝑦 ∈ 𝐷) → 𝐴 ∈ 𝐷)
21ex 418 . . . . . . . . . 10 (𝜑 → (𝑦 ∈ 𝐷 → 𝐴 ∈ 𝐷))
3 ibibr 371 . . . . . . . . . 10 ((𝑦 ∈ 𝐷 → 𝐴 ∈ 𝐷) ↔ (𝑦 ∈ 𝐷 → (𝐴 ∈ 𝐷 ↔ 𝑦 ∈ 𝐷)))
42, 3sylib 221 . . . . . . . . 9 (𝜑 → (𝑦 ∈ 𝐷 → (𝐴 ∈ 𝐷 ↔ 𝑦 ∈ 𝐷)))
54imp 412 . . . . . . . 8 ((𝜑 ∧ 𝑦 ∈ 𝐷) → (𝐴 ∈ 𝐷 ↔ 𝑦 ∈ 𝐷))
65anbi1d 643 . . . . . . 7 ((𝜑 ∧ 𝑦 ∈ 𝐷) → ((𝐴 ∈ 𝐷 ∧ 𝜒) ↔ (𝑦 ∈ 𝐷 ∧ 𝜒)))
7 rabxfrd.4 . . . . . . . 8 (𝑥 = 𝐴 → (𝜓 ↔ 𝜒))
87elrab 3644 . . . . . . 7 (𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ (𝐴 ∈ 𝐷 ∧ 𝜒))
9 rabid 3432 . . . . . . 7 (𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒} ↔ (𝑦 ∈ 𝐷 ∧ 𝜒))
106, 8, 93bitr4g 317 . . . . . 6 ((𝜑 ∧ 𝑦 ∈ 𝐷) → (𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}))
1110rabbidva 3418 . . . . 5 (𝜑 → {𝑦 ∈ 𝐷 ∣ 𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}} = {𝑦 ∈ 𝐷 ∣ 𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}})
1211eleq2d 2846 . . . 4 (𝜑 → (𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}}))
13 rabxfrd.1 . . . . 5 Ⅎ𝑦𝐵
14 nfcv 2922 . . . . 5 Ⅎ𝑦𝐷
15 rabxfrd.2 . . . . . 6 Ⅎ𝑦𝐶
1615nfel1 2938 . . . . 5 Ⅎ𝑦 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}
17 rabxfrd.5 . . . . . 6 (𝑦 = 𝐵 → 𝐴 = 𝐶)
1817eleq1d 2845 . . . . 5 (𝑦 = 𝐵 → (𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}))
1913, 14, 16, 18elrabf 3641 . . . 4 (𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}} ↔ (𝐵 ∈ 𝐷 ∧ 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}))
20 nfrab1 3431 . . . . . 6 Ⅎ𝑦{𝑦 ∈ 𝐷 ∣ 𝜒}
2113, 20nfel 2936 . . . . 5 Ⅎ𝑦 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}
22 eleq1 2848 . . . . 5 (𝑦 = 𝐵 → (𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}))
2313, 14, 21, 22elrabf 3641 . . . 4 (𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}} ↔ (𝐵 ∈ 𝐷 ∧ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}))
2412, 19, 233bitr3g 316 . . 3 (𝜑 → ((𝐵 ∈ 𝐷 ∧ 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}) ↔ (𝐵 ∈ 𝐷 ∧ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})))
25 pm5.32 584 . . 3 ((𝐵 ∈ 𝐷 → (𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})) ↔ ((𝐵 ∈ 𝐷 ∧ 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}) ↔ (𝐵 ∈ 𝐷 ∧ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})))
2624, 25sylibr 237 . 2 (𝜑 → (𝐵 ∈ 𝐷 → (𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})))
2726imp 412 1 ((𝜑 ∧ 𝐵 ∈ 𝐷) → (𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  Ⅎwnfc 2907  {crab 3412
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-tru 1573  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
This theorem is used by:  rabxfr  5379  riotaxfrd  7399
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