Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
Mirrors > Home > MPE Home > Th. List > sbhypf | Structured version Visualization version GIF version |
Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf 3865. (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 | eqeq1 2743 | . . 3 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴)) | |
2 | 1 | equsexvw 2011 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) ↔ 𝑦 = 𝐴) |
3 | nfs1v 2156 | . . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑 | |
4 | sbhypf.1 | . . . 4 ⊢ Ⅎ𝑥𝜓 | |
5 | 3, 4 | nfbi 1909 | . . 3 ⊢ Ⅎ𝑥([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
6 | sbequ12 2247 | . . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑦 / 𝑥]𝜑)) | |
7 | 6 | bicomd 222 | . . . 4 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑)) |
8 | sbhypf.2 | . . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) | |
9 | 7, 8 | sylan9bb 509 | . . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
10 | 5, 9 | exlimi 2213 | . 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
11 | 2, 10 | sylbir 234 | 1 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ 𝜓)) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 205 ∧ wa 395 = wceq 1541 ∃wex 1785 Ⅎwnf 1789 [wsb 2070 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1801 ax-4 1815 ax-5 1916 ax-6 1974 ax-7 2014 ax-9 2119 ax-10 2140 ax-12 2174 ax-ext 2710 |
This theorem depends on definitions: df-bi 206 df-an 396 df-or 844 df-tru 1544 df-ex 1786 df-nf 1790 df-sb 2071 df-cleq 2731 |
This theorem is referenced by: mob2 3653 reu2eqd 3674 cbvrabcsfw 3880 cbvopab1 5153 ralxpf 5752 cbviotaw 6395 cbvriotaw 7234 tfisi 7693 ac6sf 10229 nn0ind-raph 12403 ac6sf2 30939 nn0min 31113 ac6gf 35869 fdc1 35883 |
Copyright terms: Public domain | W3C validator |