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Mathbox for Thierry Arnoux |
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Mirrors > Home > MPE Home > Th. List > Mathboxes > eqrrabd | Structured version Visualization version GIF version |
Description: Deduce equality with a restricted abstraction. (Contributed by Thierry Arnoux, 11-Apr-2024.) |
Ref | Expression |
---|---|
eqrrabd.1 | ⊢ (𝜑 → 𝐵 ⊆ 𝐴) |
eqrrabd.2 | ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑥 ∈ 𝐵 ↔ 𝜓)) |
Ref | Expression |
---|---|
eqrrabd | ⊢ (𝜑 → 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜓}) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | nfv 1915 | . 2 ⊢ Ⅎ𝑥𝜑 | |
2 | nfcv 2901 | . 2 ⊢ Ⅎ𝑥𝐵 | |
3 | nfrab1 3449 | . 2 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐴 ∣ 𝜓} | |
4 | eqrrabd.1 | . . . . . 6 ⊢ (𝜑 → 𝐵 ⊆ 𝐴) | |
5 | 4 | sseld 3980 | . . . . 5 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)) |
6 | 5 | pm4.71rd 561 | . . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))) |
7 | eqrrabd.2 | . . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑥 ∈ 𝐵 ↔ 𝜓)) | |
8 | 7 | pm5.32da 577 | . . . 4 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝜓))) |
9 | 6, 8 | bitrd 278 | . . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝜓))) |
10 | rabid 3450 | . . 3 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜓} ↔ (𝑥 ∈ 𝐴 ∧ 𝜓)) | |
11 | 9, 10 | bitr4di 288 | . 2 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ 𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜓})) |
12 | 1, 2, 3, 11 | eqrd 4000 | 1 ⊢ (𝜑 → 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜓}) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 205 ∧ wa 394 = wceq 1539 ∈ wcel 2104 {crab 3430 ⊆ wss 3947 |
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 1911 ax-6 1969 ax-7 2009 ax-8 2106 ax-9 2114 ax-10 2135 ax-11 2152 ax-12 2169 ax-ext 2701 |
This theorem depends on definitions: df-bi 206 df-an 395 df-or 844 df-tru 1542 df-ex 1780 df-nf 1784 df-sb 2066 df-clab 2708 df-cleq 2722 df-clel 2808 df-nfc 2883 df-rab 3431 df-v 3474 df-in 3954 df-ss 3964 |
This theorem is referenced by: (None) |
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