Theorem had0 1605

Description: If the first input is false, then the adder sum is equivalent to the exclusive disjunction of the other two inputs. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 12-Jul-2020.)

Assertion

Ref	Expression
had0	⊢ (¬ 𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒)))

Proof of Theorem had0

Step	Hyp	Ref	Expression
1		had1 1604	. . 3 ⊢ (¬ 𝜑 → (hadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒)))
2		hadnot 1603	. . 3 ⊢ (¬ hadd(𝜑, 𝜓, 𝜒) ↔ hadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒))
3		xnor 1503	. . . 4 ⊢ ((𝜓 ↔ 𝜒) ↔ ¬ (𝜓 ⊻ 𝜒))
4		notbi 321	. . . 4 ⊢ ((𝜓 ↔ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒))
5	3, 4	bitr3i 279	. . 3 ⊢ (¬ (𝜓 ⊻ 𝜒) ↔ (¬ 𝜓 ↔ ¬ 𝜒))
6	1, 2, 5	3bitr4g 316	. 2 ⊢ (¬ 𝜑 → (¬ hadd(𝜑, 𝜓, 𝜒) ↔ ¬ (𝜓 ⊻ 𝜒)))
7	6	con4bid 319	1 ⊢ (¬ 𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ⊻ wxo 1501  haddwhad 1593

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-xor 1502  df-had 1594

This theorem is referenced by:  hadifp  1606  sadadd2lem2  15801  saddisjlem  15815