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| Mirrors > Home > MPE Home > Th. List > ifor | Structured version Visualization version GIF version | ||
| Description: Rewrite a disjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.) |
| Ref | Expression |
|---|---|
| ifor | ⊢ if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | iftrue 4531 | . . . 4 ⊢ ((𝜑 ∨ 𝜓) → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴) | |
| 2 | 1 | orcs 876 | . . 3 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴) |
| 3 | iftrue 4531 | . . 3 ⊢ (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴) | |
| 4 | 2, 3 | eqtr4d 2780 | . 2 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))) |
| 5 | iffalse 4534 | . . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵)) | |
| 6 | biorf 937 | . . . 4 ⊢ (¬ 𝜑 → (𝜓 ↔ (𝜑 ∨ 𝜓))) | |
| 7 | 6 | ifbid 4549 | . . 3 ⊢ (¬ 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∨ 𝜓), 𝐴, 𝐵)) |
| 8 | 5, 7 | eqtr2d 2778 | . 2 ⊢ (¬ 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))) |
| 9 | 4, 8 | pm2.61i 182 | 1 ⊢ if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) |
| Colors of variables: wff setvar class |
| Syntax hints: ¬ wn 3 ∨ wo 848 = wceq 1540 ifcif 4525 |
| 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-8 2110 ax-9 2118 ax-ext 2708 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-ex 1780 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-if 4526 |
| This theorem is referenced by: cantnflem1d 9728 cantnflem1 9729 |
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