Theorem cad1 1613

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		olc 864	. . . 4 ⊢ (𝜒 → (𝜑 ∨ 𝜒))
2		olc 864	. . . 4 ⊢ (𝜒 → (𝜓 ∨ 𝜒))
3	1, 2	jca 514	. . 3 ⊢ (𝜒 → ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)))
4	3	biantrud 534	. 2 ⊢ (𝜒 → ((𝜑 ∨ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)))))
5		cadan 1606	. . 3 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)))
6		3anass 1091	. . 3 ⊢ (((𝜑 ∨ 𝜓) ∧ (𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒)) ↔ ((𝜑 ∨ 𝜓) ∧ ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒))))
7	5, 6	bitri 277	. 2 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∨ 𝜓) ∧ ((𝜑 ∨ 𝜒) ∧ (𝜓 ∨ 𝜒))))
8	4, 7	syl6rbbr 292	1 ⊢ (𝜒 → (cadd(𝜑, 𝜓, 𝜒) ↔ (𝜑 ∨ 𝜓)))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398   ∨ wo 843   ∧ w3a 1083  caddwcad 1603

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-3or 1084  df-3an 1085  df-xor 1501  df-cad 1604

This theorem is referenced by:  cadifp  1615  sadadd2lem2  15793  sadcaddlem  15800