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| Mirrors > Home > ILE Home > Th. List > sbequ1 | GIF version | ||
| Description: An equality theorem for substitution. (Contributed by NM, 5-Aug-1993.) |
| Ref | Expression |
|---|---|
| sbequ1 | ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | pm3.4 333 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑)) | |
| 2 | 19.8a 1643 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) | |
| 3 | df-sb 1816 | . . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))) | |
| 4 | 1, 2, 3 | sylanbrc 421 | . 2 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑) |
| 5 | 4 | ex 115 | 1 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑)) |
| Colors of variables: wff set class |
| Syntax hints: → wi 4 ∧ wa 104 ∃wex 1545 [wsb 1815 |
| 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-gen 1502 ax-ie1 1546 ax-ie2 1547 ax-4 1563 |
| This theorem depends on definitions: df-bi 117 df-sb 1816 |
| This theorem is referenced by: sbequ12 1824 sbequi 1892 sb6rf 1906 mo2n 2114 bj-bdfindes 16958 bj-findes 16990 |
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