Theorem cad0 1618

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) ↔ (𝜑 ∧ 𝜓)))

Proof of Theorem cad0

Step	Hyp	Ref	Expression
1		df-cad 1608	. 2 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))))
2		idd 24	. . . 4 ⊢ (¬ 𝜒 → ((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜓)))
3		pm2.21 123	. . . . 5 ⊢ (¬ 𝜒 → (𝜒 → (𝜑 ∧ 𝜓)))
4	3	adantrd 494	. . . 4 ⊢ (¬ 𝜒 → ((𝜒 ∧ (𝜑 ⊻ 𝜓)) → (𝜑 ∧ 𝜓)))
5	2, 4	jaod 855	. . 3 ⊢ (¬ 𝜒 → (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))) → (𝜑 ∧ 𝜓)))
6		orc 863	. . 3 ⊢ ((𝜑 ∧ 𝜓) → ((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))))
7	5, 6	impbid1 227	. 2 ⊢ (¬ 𝜒 → (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))) ↔ (𝜑 ∧ 𝜓)))
8	1, 7	syl5bb 285	1 ⊢ (¬ 𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑 ∧ 𝜓)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398   ∨ wo 843   ⊻ wxo 1501  caddwcad 1607

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 209  df-an 399  df-or 844  df-cad 1608

This theorem is referenced by:  cadifp  1619  sadadd2lem2  15782  sadcaddlem  15789  saddisjlem  15796