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Mirrors > Home > MPE Home > Th. List > sblbis | Structured version Visualization version GIF version |
Description: Introduce left biconditional inside of a substitution. (Contributed by NM, 19-Aug-1993.) |
Ref | Expression |
---|---|
sblbis.1 | ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
Ref | Expression |
---|---|
sblbis | ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sbbi 2548 | . 2 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑)) | |
2 | sblbis.1 | . . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) | |
3 | 2 | bibi2i 326 | . 2 ⊢ (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓)) |
4 | 1, 3 | bitri 264 | 1 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 196 [wsb 2049 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-10 2174 ax-12 2203 ax-13 2408 |
This theorem depends on definitions: df-bi 197 df-an 383 df-or 837 df-ex 1853 df-nf 1858 df-sb 2050 |
This theorem is referenced by: sbie 2555 sb8eu 2652 sb8iota 6001 wl-sb8eut 33693 |
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