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Mirrors > Home > ILE Home > Th. List > ifandc | GIF version |
Description: Rewrite a conjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.) |
Ref | Expression |
---|---|
ifandc | ⊢ (DECID 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-dc 836 | . 2 ⊢ (DECID 𝜑 ↔ (𝜑 ∨ ¬ 𝜑)) | |
2 | iftrue 3554 | . . . 4 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵)) | |
3 | ibar 301 | . . . . 5 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓))) | |
4 | 3 | ifbid 3570 | . . . 4 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∧ 𝜓), 𝐴, 𝐵)) |
5 | 2, 4 | eqtr2d 2223 | . . 3 ⊢ (𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
6 | simpl 109 | . . . . . 6 ⊢ ((𝜑 ∧ 𝜓) → 𝜑) | |
7 | 6 | con3i 633 | . . . . 5 ⊢ (¬ 𝜑 → ¬ (𝜑 ∧ 𝜓)) |
8 | 7 | iffalsed 3559 | . . . 4 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵) |
9 | iffalse 3557 | . . . 4 ⊢ (¬ 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵) | |
10 | 8, 9 | eqtr4d 2225 | . . 3 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
11 | 5, 10 | jaoi 717 | . 2 ⊢ ((𝜑 ∨ ¬ 𝜑) → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
12 | 1, 11 | sylbi 121 | 1 ⊢ (DECID 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 104 ∨ wo 709 DECID wdc 835 = wceq 1364 ifcif 3549 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-in1 615 ax-in2 616 ax-io 710 ax-5 1458 ax-7 1459 ax-gen 1460 ax-ie1 1504 ax-ie2 1505 ax-8 1515 ax-11 1517 ax-4 1521 ax-17 1537 ax-i9 1541 ax-ial 1545 ax-i5r 1546 ax-ext 2171 |
This theorem depends on definitions: df-bi 117 df-dc 836 df-tru 1367 df-nf 1472 df-sb 1774 df-clab 2176 df-cleq 2182 df-clel 2185 df-if 3550 |
This theorem is referenced by: isumss 11430 |
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