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Mirrors > Home > MPE Home > Th. List > ifan | Structured version Visualization version GIF version |
Description: Rewrite a conjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.) |
Ref | Expression |
---|---|
ifan | ⊢ if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | iftrue 4535 | . . 3 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵)) | |
2 | ibar 527 | . . . 4 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓))) | |
3 | 2 | ifbid 4552 | . . 3 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∧ 𝜓), 𝐴, 𝐵)) |
4 | 1, 3 | eqtr2d 2771 | . 2 ⊢ (𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
5 | simpl 481 | . . . . 5 ⊢ ((𝜑 ∧ 𝜓) → 𝜑) | |
6 | 5 | con3i 154 | . . . 4 ⊢ (¬ 𝜑 → ¬ (𝜑 ∧ 𝜓)) |
7 | 6 | iffalsed 4540 | . . 3 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵) |
8 | iffalse 4538 | . . 3 ⊢ (¬ 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵) | |
9 | 7, 8 | eqtr4d 2773 | . 2 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
10 | 4, 9 | pm2.61i 182 | 1 ⊢ if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 ∧ wa 394 = wceq 1539 ifcif 4529 |
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 1911 ax-6 1969 ax-7 2009 ax-8 2106 ax-9 2114 ax-ext 2701 |
This theorem depends on definitions: df-bi 206 df-an 395 df-or 844 df-ex 1780 df-sb 2066 df-clab 2708 df-cleq 2722 df-clel 2808 df-if 4530 |
This theorem is referenced by: itg0 25531 iblre 25545 itgreval 25548 iblss 25556 iblss2 25557 itgle 25561 itgss 25563 itgeqa 25565 iblconst 25569 itgconst 25570 ibladdlem 25571 iblabslem 25579 iblabsr 25581 iblmulc2 25582 itgsplit 25587 bddiblnc 25593 itgcn 25596 mrsubrn 34800 itg2gt0cn 36848 ibladdnclem 36849 iblabsnclem 36856 iblmulc2nc 36858 selvvvval 41461 iblsplit 44982 |
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