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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 2317 | . 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒)) | |
2 | sbrbis.1 | . . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) | |
3 | 2 | bibi1i 341 | . 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒)) |
4 | 1, 3 | bitri 277 | 1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 208 [wsb 2069 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-10 2145 ax-12 2177 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-ex 1781 df-nf 1785 df-sb 2070 |
This theorem is referenced by: sbrbif 2321 sbabel 3017 |
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