Theorem had0 1634

Description: If the first input is false, then the adder sum is equivalent to the exclusive disjunction of the other two inputs, and conversely. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 12-Jul-2020.) Strengthen to a biconditional. (Revised by BJ, 10-Aug-2026.)

Assertion

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

Proof of Theorem had0

Step	Hyp	Ref	Expression
1		hadrot 1631	. . . 4 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ hadd(𝜓, 𝜒, 𝜑))
2		df-had 1624	. . . . 5 ⊢ (hadd(𝜓, 𝜒, 𝜑) ↔ ((𝜓 ⊻ 𝜒) ⊻ 𝜑))
3		df-xor 1542	. . . . . 6 ⊢ (((𝜓 ⊻ 𝜒) ⊻ 𝜑) ↔ ¬ ((𝜓 ⊻ 𝜒) ↔ 𝜑))
4		xor3 385	. . . . . 6 ⊢ (¬ ((𝜓 ⊻ 𝜒) ↔ 𝜑) ↔ ((𝜓 ⊻ 𝜒) ↔ ¬ 𝜑))
5	3, 4	bitri 278	. . . . 5 ⊢ (((𝜓 ⊻ 𝜒) ⊻ 𝜑) ↔ ((𝜓 ⊻ 𝜒) ↔ ¬ 𝜑))
6	2, 5	bitri 278	. . . 4 ⊢ (hadd(𝜓, 𝜒, 𝜑) ↔ ((𝜓 ⊻ 𝜒) ↔ ¬ 𝜑))
7	1, 6	bitri 278	. . 3 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜓 ⊻ 𝜒) ↔ ¬ 𝜑))
8		biass 388	. . 3 ⊢ (((hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒)) ↔ ¬ 𝜑) ↔ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜓 ⊻ 𝜒) ↔ ¬ 𝜑)))
9	7, 8	mpbir 234	. 2 ⊢ ((hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒)) ↔ ¬ 𝜑)
10	9	bicomi 227	1 ⊢ (¬ 𝜑 ↔ (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ⊻ 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ⊻ wxo 1541  haddwhad 1623

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-xor 1542  df-had 1624

This theorem is used by:  hadifp  1637  sadadd2lem2  16532  saddisjlem  16546