Theorem cad1 1619

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) ↔ (𝜑 ∨ 𝜓)))

Proof of Theorem cad1

Step	Hyp	Ref	Expression
1		cadan 1611	. . 3 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)))
2		3anass 1094	. . 3 ⊢ (((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ∧ ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒))))
3	1, 2	bitri 274	. 2 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∨ 𝜓) ∧ ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒))))
4		olc 865	. . . 4 ⊢ (𝜒 → (𝜑 ∨ 𝜒))
5		olc 865	. . . 4 ⊢ (𝜒 → (𝜓 ∨ 𝜒))
6	4, 5	jca 512	. . 3 ⊢ (𝜒 → ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)))
7	6	biantrud 532	. 2 ⊢ (𝜒 → ((𝜑 ∨ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)))))
8	3, 7	bitr4id 290	1 ⊢ (𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑 ∨ 𝜓)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396   ∨ wo 844   ∧ w3a 1086  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 206  df-an 397  df-or 845  df-3or 1087  df-3an 1088  df-xor 1507  df-cad 1609

This theorem is referenced by:  cadifp  1622  sadadd2lem2  16157  sadcaddlem  16164