Theorem had1 1632

Description: If the first input is true, then the adder sum is equivalent to the biconditionality of the other two inputs, and conversely. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 11-Jul-2020.) Strengthen to a biconditional. (Revised by BJ, 10-Aug-2026.)

Assertion

Ref	Expression
had1	⊢ (𝜑 ↔ (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ↔ 𝜒)))

Proof of Theorem had1

Step	Hyp	Ref	Expression
1		hadrot 1630	. . 3 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ hadd(𝜓, 𝜒, 𝜑))
2		hadbi 1627	. . 3 ⊢ (hadd(𝜓, 𝜒, 𝜑) ↔ ((𝜓 ↔ 𝜒) ↔ 𝜑))
3	1, 2	bitri 278	. 2 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜓 ↔ 𝜒) ↔ 𝜑))
4		birot 389	. 2 ⊢ ((𝜑 ↔ (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ↔ 𝜒))) ↔ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜓 ↔ 𝜒) ↔ 𝜑)))
5	3, 4	mpbir 234	1 ⊢ (𝜑 ↔ (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ↔ 𝜒)))

This proof depends on syntax axioms:   ↔ wb 209  haddwhad 1622

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 1541  df-had 1623

This theorem is used by:  hadifp  1636  sadadd2lem2  16620