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Mirrors > Home > ILE Home > Th. List > equsb3lem | GIF version |
Description: Lemma for equsb3 1939. (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 1514 | . 2 ⊢ (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧) | |
2 | equequ1 1700 | . 2 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝑧 ↔ 𝑦 = 𝑧)) | |
3 | 1, 2 | sbieh 1778 | 1 ⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 104 = wceq 1343 [wsb 1750 |
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 1435 ax-gen 1437 ax-ie1 1481 ax-ie2 1482 ax-8 1492 ax-4 1498 ax-17 1514 ax-i9 1518 ax-ial 1522 |
This theorem depends on definitions: df-bi 116 df-sb 1751 |
This theorem is referenced by: equsb3 1939 |
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