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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 835 | . 2 ⊢ (DECID 𝜑 ↔ (𝜑 ∨ ¬ 𝜑)) | |
2 | iftrue 3539 | . . . 4 ⊢ (𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, 𝐴, 𝐵)) | |
3 | ibar 301 | . . . . 5 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓))) | |
4 | 3 | ifbid 3555 | . . . 4 ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∧ 𝜓), 𝐴, 𝐵)) |
5 | 2, 4 | eqtr2d 2211 | . . 3 ⊢ (𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
6 | simpl 109 | . . . . . 6 ⊢ ((𝜑 ∧ 𝜓) → 𝜑) | |
7 | 6 | con3i 632 | . . . . 5 ⊢ (¬ 𝜑 → ¬ (𝜑 ∧ 𝜓)) |
8 | 7 | iffalsed 3544 | . . . 4 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = 𝐵) |
9 | iffalse 3542 | . . . 4 ⊢ (¬ 𝜑 → if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = 𝐵) | |
10 | 8, 9 | eqtr4d 2213 | . . 3 ⊢ (¬ 𝜑 → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)) |
11 | 5, 10 | jaoi 716 | . 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 708 DECID wdc 834 = wceq 1353 ifcif 3534 |
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 614 ax-in2 615 ax-io 709 ax-5 1447 ax-7 1448 ax-gen 1449 ax-ie1 1493 ax-ie2 1494 ax-8 1504 ax-11 1506 ax-4 1510 ax-17 1526 ax-i9 1530 ax-ial 1534 ax-i5r 1535 ax-ext 2159 |
This theorem depends on definitions: df-bi 117 df-dc 835 df-tru 1356 df-nf 1461 df-sb 1763 df-clab 2164 df-cleq 2170 df-clel 2173 df-if 3535 |
This theorem is referenced by: isumss 11370 |
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