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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 2733 | . . 3 ⊢ 𝑦 ∈ V | |
2 | eqeq1 2177 | . . 3 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴)) | |
3 | 1, 2 | ceqsexv 2769 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) ↔ 𝑦 = 𝐴) |
4 | nfs1v 1932 | . . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑 | |
5 | sbhypf.1 | . . . 4 ⊢ Ⅎ𝑥𝜓 | |
6 | 4, 5 | nfbi 1582 | . . 3 ⊢ Ⅎ𝑥([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
7 | sbequ12 1764 | . . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑦 / 𝑥]𝜑)) | |
8 | 7 | bicomd 140 | . . . 4 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑)) |
9 | sbhypf.2 | . . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) | |
10 | 8, 9 | sylan9bb 459 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
11 | 6, 10 | exlimi 1587 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
12 | 3, 11 | sylbir 134 | 1 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 103 ↔ wb 104 = wceq 1348 Ⅎwnf 1453 ∃wex 1485 [wsb 1755 |
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 1440 ax-gen 1442 ax-ie1 1486 ax-ie2 1487 ax-8 1497 ax-11 1499 ax-4 1503 ax-17 1519 ax-i9 1523 ax-ial 1527 ax-i5r 1528 ax-ext 2152 |
This theorem depends on definitions: df-bi 116 df-tru 1351 df-nf 1454 df-sb 1756 df-clab 2157 df-cleq 2163 df-clel 2166 df-v 2732 |
This theorem is referenced by: mob2 2910 cbvmptf 4083 tfisi 4571 ralxpf 4757 rexxpf 4758 nn0ind-raph 9329 |
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