MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  cad0 Structured version   Visualization version   GIF version

Theorem cad0 1609
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 1599 . 2 (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))))
2 idd 24 . . . 4 𝜒 → ((𝜑𝜓) → (𝜑𝜓)))
3 pm2.21 123 . . . . 5 𝜒 → (𝜒 → (𝜑𝜓)))
43adantrd 492 . . . 4 𝜒 → ((𝜒 ∧ (𝜑𝜓)) → (𝜑𝜓)))
52, 4jaod 853 . . 3 𝜒 → (((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))) → (𝜑𝜓)))
6 orc 861 . . 3 ((𝜑𝜓) → ((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))))
75, 6impbid1 226 . 2 𝜒 → (((𝜑𝜓) ∨ (𝜒 ∧ (𝜑𝜓))) ↔ (𝜑𝜓)))
81, 7syl5bb 284 1 𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑𝜓)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 207  wa 396  wo 841  wxo 1495  caddwcad 1598
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 208  df-an 397  df-or 842  df-cad 1599
This theorem is referenced by:  cadifp  1610  sadadd2lem2  15787  sadcaddlem  15794  saddisjlem  15801
  Copyright terms: Public domain W3C validator