Theorem cad0 1651

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.) (Proof shortened by Wolf Lammen, 21-Sep-2024.)

Assertion

Ref	Expression
cad0	⊢ (¬ 𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑 ∧ 𝜓)))

Proof of Theorem cad0

Step	Hyp	Ref	Expression
1		df-cad 1640	. . 3 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))))
2		idd 25	. . . 4 ⊢ (¬ 𝜒 → ((𝜑 ∧ 𝜓) → (𝜑 ∧ 𝜓)))
3		pm2.21 124	. . . . 5 ⊢ (¬ 𝜒 → (𝜒 → (𝜑 ∧ 𝜓)))
4	3	adantrd 497	. . . 4 ⊢ (¬ 𝜒 → ((𝜒 ∧ (𝜑 ⊻ 𝜓)) → (𝜑 ∧ 𝜓)))
5	2, 4	jaod 873	. . 3 ⊢ (¬ 𝜒 → (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))) → (𝜑 ∧ 𝜓)))
6	1, 5	biimtrid 245	. 2 ⊢ (¬ 𝜒 → (cadd(𝜑, 𝜓, 𝜒) → (𝜑 ∧ 𝜓)))
7		cad11 1649	. 2 ⊢ ((𝜑 ∧ 𝜓) → cadd(𝜑, 𝜓, 𝜒))
8	6, 7	impbid1 228	1 ⊢ (¬ 𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑 ∧ 𝜓)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   ⊻ wxo 1541  caddwcad 1639

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-cad 1640

This theorem is used by:  cadifp  1652  sadadd2lem2  16533  sadcaddlem  16540  saddisjlem  16547