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 Description: If one input is false, then the adder carry is true exactly when both of the other two inputs are true. (Contributed by Mario Carneiro, 8-Sep-2016.)
Assertion
Ref Expression
cad0 𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑𝜓)))

Proof of Theorem cad0
StepHypRef Expression
1 df-cad 1537 . 2 (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))))
2 idd 24 . . . 4 𝜒 → ((𝜑𝜓) → (𝜑𝜓)))
3 pm2.21 119 . . . . 5 𝜒 → (𝜒 → (𝜑𝜓)))
43adantrd 483 . . . 4 𝜒 → ((𝜒 ∧ (𝜑𝜓)) → (𝜑𝜓)))
52, 4jaod 394 . . 3 𝜒 → (((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))) → (𝜑𝜓)))
6 orc 399 . . 3 ((𝜑𝜓) → ((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))))
75, 6impbid1 214 . 2 𝜒 → (((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))) ↔ (𝜑𝜓)))
81, 7syl5bb 271 1 𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑𝜓)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∨ wo 382   ∧ wa 383   ⊻ wxo 1456  caddwcad 1536 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 196  df-or 384  df-an 385  df-cad 1537 This theorem is referenced by:  cadifp  1548  sadadd2lem2  15010  sadcaddlem  15017  saddisjlem  15024
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