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Mirrors > Home > ILE Home > Th. List > equsb3lem | GIF version |
Description: Lemma for equsb3 1963. (Contributed by NM, 4-Dec-2005.) (Proof shortened by Andrew Salmon, 14-Jun-2011.) |
Ref | Expression |
---|---|
equsb3lem | ⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | ax-17 1537 | . 2 ⊢ (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧) | |
2 | equequ1 1723 | . 2 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝑧 ↔ 𝑦 = 𝑧)) | |
3 | 1, 2 | sbieh 1801 | 1 ⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧) |
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-8 1515 ax-4 1521 ax-17 1537 ax-i9 1541 ax-ial 1545 |
This theorem depends on definitions: df-bi 117 df-sb 1774 |
This theorem is referenced by: equsb3 1963 |
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