Metamath Proof Explorer


Theorem had0

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)

Ref Expression
Assertion had0 ( ¬ 𝜑 → ( hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( 𝜓𝜒 ) ) )

Proof

Step Hyp Ref Expression
1 had1 ( ¬ 𝜑 → ( hadd ( ¬ 𝜑 , ¬ 𝜓 , ¬ 𝜒 ) ↔ ( ¬ 𝜓 ↔ ¬ 𝜒 ) ) )
2 hadnot ( ¬ hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ hadd ( ¬ 𝜑 , ¬ 𝜓 , ¬ 𝜒 ) )
3 xnor ( ( 𝜓𝜒 ) ↔ ¬ ( 𝜓𝜒 ) )
4 notbi ( ( 𝜓𝜒 ) ↔ ( ¬ 𝜓 ↔ ¬ 𝜒 ) )
5 3 4 bitr3i ( ¬ ( 𝜓𝜒 ) ↔ ( ¬ 𝜓 ↔ ¬ 𝜒 ) )
6 1 2 5 3bitr4g ( ¬ 𝜑 → ( ¬ hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ ¬ ( 𝜓𝜒 ) ) )
7 6 con4bid ( ¬ 𝜑 → ( hadd ( 𝜑 , 𝜓 , 𝜒 ) ↔ ( 𝜓𝜒 ) ) )