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Mirrors > Home > MPE Home > Th. List > had0 | Structured version Visualization version GIF version |
Description: If the first input is false, then the adder sum is equivalent to the exclusive disjunction of the other two inputs. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 12-Jul-2020.) |
Ref | Expression |
---|---|
had0 | ⊢ (¬ 𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | had1 1605 | . . 3 ⊢ (¬ 𝜑 → (hadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒))) | |
2 | hadnot 1604 | . . 3 ⊢ (¬ hadd(𝜑, 𝜓, 𝜒) ↔ hadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒)) | |
3 | xnor 1508 | . . . 4 ⊢ ((𝜓 ↔ 𝜒) ↔ ¬ (𝜓 ⊻ 𝜒)) | |
4 | notbi 319 | . . . 4 ⊢ ((𝜓 ↔ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒)) | |
5 | 3, 4 | bitr3i 276 | . . 3 ⊢ (¬ (𝜓 ⊻ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒)) |
6 | 1, 2, 5 | 3bitr4g 314 | . 2 ⊢ (¬ 𝜑 → (¬ hadd(𝜑, 𝜓, 𝜒) ↔ ¬ (𝜓 ⊻ 𝜒))) |
7 | 6 | con4bid 317 | 1 ⊢ (¬ 𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒))) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ↔ wb 205 ⊻ wxo 1506 haddwhad 1594 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 |
This theorem depends on definitions: df-bi 206 df-xor 1507 df-had 1595 |
This theorem is referenced by: hadifp 1607 sadadd2lem2 16157 saddisjlem 16171 |
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