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Theorem cad0 1546
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 1536 . 2 (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))))
2 idd 24 . . . 4 𝜒 → ((𝜑𝜓) → (𝜑𝜓)))
3 pm2.21 118 . . . . 5 𝜒 → (𝜒 → (𝜑𝜓)))
43adantrd 482 . . . 4 𝜒 → ((𝜒 ∧ (𝜑𝜓)) → (𝜑𝜓)))
52, 4jaod 393 . . 3 𝜒 → (((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))) → (𝜑𝜓)))
6 orc 398 . . 3 ((𝜑𝜓) → ((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))))
75, 6impbid1 213 . 2 𝜒 → (((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))) ↔ (𝜑𝜓)))
81, 7syl5bb 270 1 𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑𝜓)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 194  wo 381  wa 382  wxo 1455  caddwcad 1535
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 195  df-or 383  df-an 384  df-cad 1536
This theorem is referenced by:  cadifp  1547  sadadd2lem2  14956  sadcaddlem  14963  saddisjlem  14970
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