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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 2740 | . . 3 ⊢ 𝑦 ∈ V | |
2 | eqeq1 2184 | . . 3 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴)) | |
3 | 1, 2 | ceqsexv 2776 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) ↔ 𝑦 = 𝐴) |
4 | nfs1v 1939 | . . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑 | |
5 | sbhypf.1 | . . . 4 ⊢ Ⅎ𝑥𝜓 | |
6 | 4, 5 | nfbi 1589 | . . 3 ⊢ Ⅎ𝑥([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
7 | sbequ12 1771 | . . . . 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 1594 | . 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 1353 Ⅎwnf 1460 ∃wex 1492 [wsb 1762 |
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 1447 ax-gen 1449 ax-ie1 1493 ax-ie2 1494 ax-8 1504 ax-11 1506 ax-4 1510 ax-17 1526 ax-i9 1530 ax-ial 1534 ax-i5r 1535 ax-ext 2159 |
This theorem depends on definitions: df-bi 117 df-tru 1356 df-nf 1461 df-sb 1763 df-clab 2164 df-cleq 2170 df-clel 2173 df-v 2739 |
This theorem is referenced by: mob2 2917 cbvmptf 4097 tfisi 4586 ralxpf 4773 rexxpf 4774 nn0ind-raph 9369 |
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