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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 2305 | . 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒)) | |
2 | sbrbis.1 | . . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) | |
3 | 2 | bibi1i 339 | . 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 205 [wsb 2068 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1914 ax-6 1972 ax-7 2012 ax-10 2138 ax-12 2172 |
This theorem depends on definitions: df-bi 206 df-an 398 df-or 847 df-ex 1783 df-nf 1787 df-sb 2069 |
This theorem is referenced by: sbrbif 2308 sbabel 2942 sbabelOLD 2943 |
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