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Mirrors > Home > ILE Home > Th. List > Mathboxes > elabgf0 | GIF version |
Description: Lemma for elabgf 2800. (Contributed by BJ, 21-Nov-2019.) |
Ref | Expression |
---|---|
elabgf0 | ⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | abid 2105 | . 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑) | |
2 | eleq1 2180 | . 2 ⊢ (𝑥 = 𝐴 → (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})) | |
3 | 1, 2 | syl5rbbr 194 | 1 ⊢ (𝑥 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 104 = wceq 1316 ∈ wcel 1465 {cab 2103 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-5 1408 ax-gen 1410 ax-ie1 1454 ax-ie2 1455 ax-8 1467 ax-4 1472 ax-17 1491 ax-i9 1495 ax-ial 1499 ax-ext 2099 |
This theorem depends on definitions: df-bi 116 df-sb 1721 df-clab 2104 df-cleq 2110 df-clel 2113 |
This theorem is referenced by: elabgft1 12912 elabgf2 12914 |
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