Theorem cador 1640

Description: The adder carry in disjunctive normal form. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 11-Jul-2020.)

Assertion

Ref	Expression
cador	⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∧ 𝜓) ∨ (𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒)))

Proof of Theorem cador

Step	Hyp	Ref	Expression
1		xor2 1546	. . . . . . 7 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)))
2	1	rbaib 548	. . . . . 6 ⊢ (¬ (𝜑 ∧ 𝜓) → ((𝜑 ⊻ 𝜓) ↔ (𝜑 ∨ 𝜓)))
3	2	anbi1d 643	. . . . 5 ⊢ (¬ (𝜑 ∧ 𝜓) → (((𝜑 ⊻ 𝜓) ∧ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∧ 𝜒)))
4		ancom 466	. . . . 5 ⊢ (((𝜑 ⊻ 𝜓) ∧ 𝜒) ↔ (𝜒 ∧ (𝜑 ⊻ 𝜓)))
5		andir 1026	. . . . 5 ⊢ (((𝜑 ∨ 𝜓) ∧ 𝜒) ↔ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒)))
6	3, 4, 5	3bitr3g 316	. . . 4 ⊢ (¬ (𝜑 ∧ 𝜓) → ((𝜒 ∧ (𝜑 ⊻ 𝜓)) ↔ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
7	6	pm5.74i 274	. . 3 ⊢ ((¬ (𝜑 ∧ 𝜓) → (𝜒 ∧ (𝜑 ⊻ 𝜓))) ↔ (¬ (𝜑 ∧ 𝜓) → ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
8		df-or 862	. . 3 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))) ↔ (¬ (𝜑 ∧ 𝜓) → (𝜒 ∧ (𝜑 ⊻ 𝜓))))
9		df-or 862	. . 3 ⊢ (((𝜑 ∧ 𝜓) ∨ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))) ↔ (¬ (𝜑 ∧ 𝜓) → ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
10	7, 8, 9	3bitr4i 306	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))) ↔ ((𝜑 ∧ 𝜓) ∨ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
11		df-cad 1639	. 2 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))))
12		3orass 1106	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ∨ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
13	10, 11, 12	3bitr4i 306	1 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∧ 𝜓) ∨ (𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   ∨ w3o 1102   ⊻ wxo 1540  caddwcad 1638

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-3or 1104  df-xor 1541  df-cad 1639

This theorem is used by:  cadan  1641  cadnot  1647