Theorem had0 1403

Description: If the first parameter is false, the half adder is equivalent to the XOR of the other two inputs. (Contributed by Mario Carneiro, 4-Sep-2016.)

Assertion

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

Proof of Theorem had0

Step	Hyp	Ref	Expression
1		had1 1402	. . 3 ⊢ (¬ φ → (hadd(¬ φ, ¬ ψ, ¬ χ) ↔ (¬ ψ ↔ ¬ χ)))
2		hadnot 1393	. . 3 ⊢ (¬ hadd(φ, ψ, χ) ↔ hadd(¬ φ, ¬ ψ, ¬ χ))
3		df-xor 1305	. . . . 5 ⊢ ((¬ ψ ⊻ ¬ χ) ↔ ¬ (¬ ψ ↔ ¬ χ))
4		xorneg 1313	. . . . 5 ⊢ ((¬ ψ ⊻ ¬ χ) ↔ (ψ ⊻ χ))
5	3, 4	bitr3i 242	. . . 4 ⊢ (¬ (¬ ψ ↔ ¬ χ) ↔ (ψ ⊻ χ))
6	5	con1bii 321	. . 3 ⊢ (¬ (ψ ⊻ χ) ↔ (¬ ψ ↔ ¬ χ))
7	1, 2, 6	3bitr4g 279	. 2 ⊢ (¬ φ → (¬ hadd(φ, ψ, χ) ↔ ¬ (ψ ⊻ χ)))
8	7	con4bid 284	1 ⊢ (¬ φ → (hadd(φ, ψ, χ) ↔ (ψ ⊻ χ)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 176   ⊻ wxo 1304  haddwhad 1378

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-xor 1305  df-had 1380

This theorem is referenced by: (None)