Theorem cador 1610

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 1513	. . . . . . 7 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)))
2	1	rbaib 539	. . . . . 6 ⊢ (¬ (𝜑 ∧ 𝜓) → ((𝜑 ⊻ 𝜓) ↔ (𝜑 ∨ 𝜓)))
3	2	anbi1d 630	. . . . 5 ⊢ (¬ (𝜑 ∧ 𝜓) → (((𝜑 ⊻ 𝜓) ∧ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∧ 𝜒)))
4		ancom 461	. . . . 5 ⊢ (((𝜑 ⊻ 𝜓) ∧ 𝜒) ↔ (𝜒 ∧ (𝜑 ⊻ 𝜓)))
5		andir 1006	. . . . 5 ⊢ (((𝜑 ∨ 𝜓) ∧ 𝜒) ↔ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒)))
6	3, 4, 5	3bitr3g 313	. . . 4 ⊢ (¬ (𝜑 ∧ 𝜓) → ((𝜒 ∧ (𝜑 ⊻ 𝜓)) ↔ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
7	6	pm5.74i 270	. . 3 ⊢ ((¬ (𝜑 ∧ 𝜓) → (𝜒 ∧ (𝜑 ⊻ 𝜓))) ↔ (¬ (𝜑 ∧ 𝜓) → ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
8		df-or 845	. . 3 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))) ↔ (¬ (𝜑 ∧ 𝜓) → (𝜒 ∧ (𝜑 ⊻ 𝜓))))
9		df-or 845	. . 3 ⊢ (((𝜑 ∧ 𝜓) ∨ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))) ↔ (¬ (𝜑 ∧ 𝜓) → ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
10	7, 8, 9	3bitr4i 303	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))) ↔ ((𝜑 ∧ 𝜓) ∨ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
11		df-cad 1609	. 2 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ (𝜑 ⊻ 𝜓))))
12		3orass 1089	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ∨ ((𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒))))
13	10, 11, 12	3bitr4i 303	1 ⊢ (cadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ∧ 𝜓) ∨ (𝜑 ∧ 𝜒) ∨ (𝜓 ∧ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 396   ∨ wo 844   ∨ w3o 1085   ⊻ wxo 1506  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-xor 1507  df-cad 1609

This theorem is referenced by:  cadan  1611  cadnot  1617