![]() |
Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
|
Mirrors > Home > MPE Home > Th. List > sbrbis | Structured version Visualization version GIF version |
Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.) |
Ref | Expression |
---|---|
sbrbis.1 | ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
Ref | Expression |
---|---|
sbrbis | ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sbbi 2307 | . 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒)) | |
2 | sbrbis.1 | . . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) | |
3 | 2 | bibi1i 338 | . 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒)) |
4 | 1, 3 | bitri 275 | 1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 206 [wsb 2062 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-10 2139 ax-12 2175 |
This theorem depends on definitions: df-bi 207 df-an 396 df-ex 1777 df-nf 1781 df-sb 2063 |
This theorem is referenced by: sbrbif 2310 sbabel 2936 sbabelOLD 2937 |
Copyright terms: Public domain | W3C validator |