Theorem had0

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)

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

Proof

Step	Hyp	Ref	Expression
1		hadrot	⊢ ( hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ hadd ( 𝜓 , 𝜒 , 𝜑 ) )
2		df-had	⊢ ( hadd ( 𝜓 , 𝜒 , 𝜑 ) ↔ ( ( 𝜓 ⊻ 𝜒 ) ⊻ 𝜑 ) )
3		df-xor	⊢ ( ( ( 𝜓 ⊻ 𝜒 ) ⊻ 𝜑 ) ↔ ¬ ( ( 𝜓 ⊻ 𝜒 ) ↔ 𝜑 ) )
4		xor3	⊢ ( ¬ ( ( 𝜓 ⊻ 𝜒 ) ↔ 𝜑 ) ↔ ( ( 𝜓 ⊻ 𝜒 ) ↔ ¬ 𝜑 ) )
5	3 4	bitri	⊢ ( ( ( 𝜓 ⊻ 𝜒 ) ⊻ 𝜑 ) ↔ ( ( 𝜓 ⊻ 𝜒 ) ↔ ¬ 𝜑 ) )
6	2 5	bitri	⊢ ( hadd ( 𝜓 , 𝜒 , 𝜑 ) ↔ ( ( 𝜓 ⊻ 𝜒 ) ↔ ¬ 𝜑 ) )
7	1 6	bitri	⊢ ( hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( ( 𝜓 ⊻ 𝜒 ) ↔ ¬ 𝜑 ) )
8		biass	⊢ ( ( ( hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( 𝜓 ⊻ 𝜒 ) ) ↔ ¬ 𝜑 ) ↔ ( hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( ( 𝜓 ⊻ 𝜒 ) ↔ ¬ 𝜑 ) ) )
9	7 8	mpbir	⊢ ( ( hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( 𝜓 ⊻ 𝜒 ) ) ↔ ¬ 𝜑 )
10	9	bicomi	⊢ ( ¬ 𝜑 ↔ ( hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( 𝜓 ⊻ 𝜒 ) ) )