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	|- ( -. ph <-> ( hadd ( ph , ps , ch ) <-> ( ps \/_ ch ) ) )

Proof

Step	Hyp	Ref	Expression
1		hadrot	|- ( hadd ( ph , ps , ch ) <-> hadd ( ps , ch , ph ) )
2		df-had	|- ( hadd ( ps , ch , ph ) <-> ( ( ps \/_ ch ) \/_ ph ) )
3		df-xor	|- ( ( ( ps \/_ ch ) \/_ ph ) <-> -. ( ( ps \/_ ch ) <-> ph ) )
4		xor3	|- ( -. ( ( ps \/_ ch ) <-> ph ) <-> ( ( ps \/_ ch ) <-> -. ph ) )
5	3 4	bitri	|- ( ( ( ps \/_ ch ) \/_ ph ) <-> ( ( ps \/_ ch ) <-> -. ph ) )
6	2 5	bitri	|- ( hadd ( ps , ch , ph ) <-> ( ( ps \/_ ch ) <-> -. ph ) )
7	1 6	bitri	|- ( hadd ( ph , ps , ch ) <-> ( ( ps \/_ ch ) <-> -. ph ) )
8		biass	|- ( ( ( hadd ( ph , ps , ch ) <-> ( ps \/_ ch ) ) <-> -. ph ) <-> ( hadd ( ph , ps , ch ) <-> ( ( ps \/_ ch ) <-> -. ph ) ) )
9	7 8	mpbir	|- ( ( hadd ( ph , ps , ch ) <-> ( ps \/_ ch ) ) <-> -. ph )
10	9	bicomi	|- ( -. ph <-> ( hadd ( ph , ps , ch ) <-> ( ps \/_ ch ) ) )