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| 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 2308 | . 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 2064 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-10 2141 ax-12 2177 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-ex 1780 df-nf 1784 df-sb 2065 | 
| This theorem is referenced by: sbrbif 2311 sbabel 2938 sbabelOLD 2939 | 
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