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Theorem had0 1541
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.)
Assertion
Ref Expression
had0 𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓𝜒)))

Proof of Theorem had0
StepHypRef Expression
1 had1 1540 . . 3 𝜑 → (hadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒)))
2 hadnot 1539 . . 3 (¬ hadd(𝜑, 𝜓, 𝜒) ↔ hadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒))
3 xnor 1464 . . . 4 ((𝜓𝜒) ↔ ¬ (𝜓𝜒))
4 notbi 309 . . . 4 ((𝜓𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒))
53, 4bitr3i 266 . . 3 (¬ (𝜓𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒))
61, 2, 53bitr4g 303 . 2 𝜑 → (¬ hadd(𝜑, 𝜓, 𝜒) ↔ ¬ (𝜓𝜒)))
76con4bid 307 1 𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓𝜒)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 196  wxo 1462  haddwhad 1530
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 197  df-xor 1463  df-had 1531
This theorem is referenced by:  hadifp  1542  sadadd2lem2  15153  saddisjlem  15167
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