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| Mirrors > Home > ILE Home > Th. List > sbhypf | GIF version | ||
| Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf . (Contributed by Raph Levien, 10-Apr-2004.) |
| Ref | Expression |
|---|---|
| sbhypf.1 | ⊢ Ⅎ𝑥𝜓 |
| sbhypf.2 | ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) |
| Ref | Expression |
|---|---|
| sbhypf | ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | vex 2766 | . . 3 ⊢ 𝑦 ∈ V | |
| 2 | eqeq1 2203 | . . 3 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴)) | |
| 3 | 1, 2 | ceqsexv 2802 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) ↔ 𝑦 = 𝐴) |
| 4 | nfs1v 1958 | . . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑 | |
| 5 | sbhypf.1 | . . . 4 ⊢ Ⅎ𝑥𝜓 | |
| 6 | 4, 5 | nfbi 1603 | . . 3 ⊢ Ⅎ𝑥([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
| 7 | sbequ12 1785 | . . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑦 / 𝑥]𝜑)) | |
| 8 | 7 | bicomd 141 | . . . 4 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑)) |
| 9 | sbhypf.2 | . . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) | |
| 10 | 8, 9 | sylan9bb 462 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
| 11 | 6, 10 | exlimi 1608 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
| 12 | 3, 11 | sylbir 135 | 1 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
| Colors of variables: wff set class |
| Syntax hints: → wi 4 ∧ wa 104 ↔ wb 105 = wceq 1364 Ⅎwnf 1474 ∃wex 1506 [wsb 1776 |
| 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 1461 ax-gen 1463 ax-ie1 1507 ax-ie2 1508 ax-8 1518 ax-11 1520 ax-4 1524 ax-17 1540 ax-i9 1544 ax-ial 1548 ax-i5r 1549 ax-ext 2178 |
| This theorem depends on definitions: df-bi 117 df-tru 1367 df-nf 1475 df-sb 1777 df-clab 2183 df-cleq 2189 df-clel 2192 df-v 2765 |
| This theorem is referenced by: mob2 2944 cbvmptf 4127 tfisi 4623 ralxpf 4812 rexxpf 4813 nn0ind-raph 9443 |
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