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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 2604 | . . 3 ⊢ 𝑦 ∈ V | |
2 | eqeq1 2087 | . . 3 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴)) | |
3 | 1, 2 | ceqsexv 2638 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) ↔ 𝑦 = 𝐴) |
4 | nfs1v 1856 | . . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑 | |
5 | sbhypf.1 | . . . 4 ⊢ Ⅎ𝑥𝜓 | |
6 | 4, 5 | nfbi 1521 | . . 3 ⊢ Ⅎ𝑥([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
7 | sbequ12 1694 | . . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑦 / 𝑥]𝜑)) | |
8 | 7 | bicomd 139 | . . . 4 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑)) |
9 | sbhypf.2 | . . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) | |
10 | 8, 9 | sylan9bb 449 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
11 | 6, 10 | exlimi 1525 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
12 | 3, 11 | sylbir 133 | 1 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 102 ↔ wb 103 = wceq 1284 Ⅎwnf 1389 ∃wex 1421 [wsb 1685 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-5 1376 ax-gen 1378 ax-ie1 1422 ax-ie2 1423 ax-8 1435 ax-11 1437 ax-4 1440 ax-17 1459 ax-i9 1463 ax-ial 1467 ax-i5r 1468 ax-ext 2063 |
This theorem depends on definitions: df-bi 115 df-tru 1287 df-nf 1390 df-sb 1686 df-clab 2068 df-cleq 2074 df-clel 2077 df-v 2603 |
This theorem is referenced by: mob2 2772 tfisi 4328 ralxpf 4500 rexxpf 4501 nn0ind-raph 8464 |
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