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Mirrors > Home > ILE Home > Th. List > eqsb1lem | GIF version |
Description: Lemma for eqsb1 2297. (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 14-Jun-2011.) |
Ref | Expression |
---|---|
eqsb1lem | ⊢ ([𝑦 / 𝑥]𝑥 = 𝐴 ↔ 𝑦 = 𝐴) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | nfv 1539 | . 2 ⊢ Ⅎ𝑥 𝑦 = 𝐴 | |
2 | eqeq1 2200 | . 2 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴)) | |
3 | 1, 2 | sbie 1802 | 1 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐴 ↔ 𝑦 = 𝐴) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 105 = wceq 1364 [wsb 1773 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-5 1458 ax-gen 1460 ax-ie1 1504 ax-ie2 1505 ax-4 1521 ax-17 1537 ax-i9 1541 ax-ial 1545 ax-ext 2175 |
This theorem depends on definitions: df-bi 117 df-nf 1472 df-sb 1774 df-cleq 2186 |
This theorem is referenced by: eqsb1 2297 |
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