![]() |
Intuitionistic Logic Explorer |
< Previous
Next >
Nearby theorems |
|
Mirrors > Home > ILE Home > Th. List > ifordc | GIF version |
Description: Rewrite a disjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014.) |
Ref | Expression |
---|---|
ifordc | ⊢ (DECID 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | exmiddc 836 | . 2 ⊢ (DECID 𝜑 → (𝜑 ∨ ¬ 𝜑)) | |
2 | iftrue 3539 | . . . . 5 ⊢ ((𝜑 ∨ 𝜓) → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴) | |
3 | 2 | orcs 735 | . . . 4 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = 𝐴) |
4 | iftrue 3539 | . . . 4 ⊢ (𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = 𝐴) | |
5 | 3, 4 | eqtr4d 2213 | . . 3 ⊢ (𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))) |
6 | iffalse 3542 | . . . 4 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵)) = if(𝜓, 𝐴, 𝐵)) | |
7 | biorf 744 | . . . . 5 ⊢ (¬ 𝜑 → (𝜓 ↔ (𝜑 ∨ 𝜓))) | |
8 | 7 | ifbid 3555 | . . . 4 ⊢ (¬ 𝜑 → if(𝜓, 𝐴, 𝐵) = if((𝜑 ∨ 𝜓), 𝐴, 𝐵)) |
9 | 6, 8 | eqtr2d 2211 | . . 3 ⊢ (¬ 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))) |
10 | 5, 9 | jaoi 716 | . 2 ⊢ ((𝜑 ∨ ¬ 𝜑) → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))) |
11 | 1, 10 | syl 14 | 1 ⊢ (DECID 𝜑 → if((𝜑 ∨ 𝜓), 𝐴, 𝐵) = if(𝜑, 𝐴, if(𝜓, 𝐴, 𝐵))) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∨ 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: nninfwlpoimlemg 7166 |
Copyright terms: Public domain | W3C validator |