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Theorem cad1 1552
 Description: If one input is true, then the adder carry is true exactly when at least one of the other two inputs is true. (Contributed by Mario Carneiro, 8-Sep-2016.) (Proof shortened by Wolf Lammen, 19-Jun-2020.)
Assertion
Ref Expression
cad1 (𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑𝜓)))

Proof of Theorem cad1
StepHypRef Expression
1 olc 399 . . . 4 (𝜒 → (𝜑𝜒))
2 olc 399 . . . 4 (𝜒 → (𝜓𝜒))
31, 2jca 554 . . 3 (𝜒 → ((𝜑𝜒) ∧ (𝜓𝜒)))
43biantrud 528 . 2 (𝜒 → ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ ((𝜑𝜒) ∧ (𝜓𝜒)))))
5 cadan 1545 . . 3 (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑𝜓) ∧ (𝜑𝜒) ∧ (𝜓𝜒)))
6 3anass 1040 . . 3 (((𝜑𝜓) ∧ (𝜑𝜒) ∧ (𝜓𝜒)) ↔ ((𝜑𝜓) ∧ ((𝜑𝜒) ∧ (𝜓𝜒))))
75, 6bitri 264 . 2 (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑𝜓) ∧ ((𝜑𝜒) ∧ (𝜓𝜒))))
84, 7syl6rbbr 279 1 (𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑𝜓)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 196   ∨ wo 383   ∧ wa 384   ∧ w3a 1036  caddwcad 1542 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-or 385  df-an 386  df-3or 1037  df-3an 1038  df-xor 1462  df-cad 1543 This theorem is referenced by:  cadifp  1554  sadadd2lem2  15096  sadcaddlem  15103
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