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Theorem cad0 1619
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 1609 . 2 (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))))
2 idd 24 . . . 4 𝜒 → ((𝜑𝜓) → (𝜑𝜓)))
3 pm2.21 123 . . . . 5 𝜒 → (𝜒 → (𝜑𝜓)))
43adantrd 495 . . . 4 𝜒 → ((𝜒 ∧ (𝜑𝜓)) → (𝜑𝜓)))
52, 4jaod 856 . . 3 𝜒 → (((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))) → (𝜑𝜓)))
6 orc 864 . . 3 ((𝜑𝜓) → ((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))))
75, 6impbid1 228 . 2 𝜒 → (((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))) ↔ (𝜑𝜓)))
81, 7syl5bb 286 1 𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑𝜓)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 209  wa 399  wo 844  wxo 1502  caddwcad 1608
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 210  df-an 400  df-or 845  df-cad 1609
This theorem is referenced by:  cadifp  1620  sadadd2lem2  15786  sadcaddlem  15793  saddisjlem  15800
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