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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 1916 | . 2 ⊢ Ⅎ𝑥𝜑 | |
2 | nfcv 2917 | . 2 ⊢ Ⅎ𝑥𝐵 | |
3 | nfrab1 3300 | . 2 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐴 ∣ 𝜓} | |
4 | eqrrabd.1 | . . . . . 6 ⊢ (𝜑 → 𝐵 ⊆ 𝐴) | |
5 | 4 | sseld 3887 | . . . . 5 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)) |
6 | 5 | pm4.71rd 567 | . . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))) |
7 | eqrrabd.2 | . . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑥 ∈ 𝐵 ↔ 𝜓)) | |
8 | 7 | pm5.32da 583 | . . . 4 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝜓))) |
9 | 6, 8 | bitrd 282 | . . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝜓))) |
10 | rabid 3294 | . . 3 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜓} ↔ (𝑥 ∈ 𝐴 ∧ 𝜓)) | |
11 | 9, 10 | bitr4di 293 | . 2 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ 𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜓})) |
12 | 1, 2, 3, 11 | eqrd 3907 | 1 ⊢ (𝜑 → 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜓}) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 209 ∧ wa 400 = wceq 1539 ∈ wcel 2112 {crab 3072 ⊆ wss 3854 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2730 |
This theorem depends on definitions: df-bi 210 df-an 401 df-or 846 df-tru 1542 df-ex 1783 df-nf 1787 df-sb 2071 df-clab 2737 df-cleq 2751 df-clel 2831 df-nfc 2899 df-rab 3077 df-v 3409 df-in 3861 df-ss 3871 |
This theorem is referenced by: (None) |
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