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Mirrors > Home > ILE Home > Th. List > eqsb3lem | GIF version |
Description: Lemma for eqsb3 2198. (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 14-Jun-2011.) |
Ref | Expression |
---|---|
eqsb3lem | ⊢ ([𝑥 / 𝑦]𝑦 = 𝐴 ↔ 𝑥 = 𝐴) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | nfv 1473 | . 2 ⊢ Ⅎ𝑦 𝑥 = 𝐴 | |
2 | eqeq1 2101 | . 2 ⊢ (𝑦 = 𝑥 → (𝑦 = 𝐴 ↔ 𝑥 = 𝐴)) | |
3 | 1, 2 | sbie 1728 | 1 ⊢ ([𝑥 / 𝑦]𝑦 = 𝐴 ↔ 𝑥 = 𝐴) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 104 = wceq 1296 [wsb 1699 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-5 1388 ax-gen 1390 ax-ie1 1434 ax-ie2 1435 ax-4 1452 ax-17 1471 ax-i9 1475 ax-ial 1479 ax-ext 2077 |
This theorem depends on definitions: df-bi 116 df-nf 1402 df-sb 1700 df-cleq 2088 |
This theorem is referenced by: eqsb3 2198 |
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