Theorem hadbi 1387

Description: The half adder is the same as the triple biconditional. (Contributed by Mario Carneiro, 4-Sep-2016.)

Assertion

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

Proof of Theorem hadbi

Step	Hyp	Ref	Expression
1		df-xor 1305	. 2 ⊢ (((φ ⊻ ψ) ⊻ χ) ↔ ¬ ((φ ⊻ ψ) ↔ χ))
2		df-had 1380	. 2 ⊢ (hadd(φ, ψ, χ) ↔ ((φ ⊻ ψ) ⊻ χ))
3		xnor 1306	. . . 4 ⊢ ((φ ↔ ψ) ↔ ¬ (φ ⊻ ψ))
4	3	bibi1i 305	. . 3 ⊢ (((φ ↔ ψ) ↔ χ) ↔ (¬ (φ ⊻ ψ) ↔ χ))
5		nbbn 347	. . 3 ⊢ ((¬ (φ ⊻ ψ) ↔ χ) ↔ ¬ ((φ ⊻ ψ) ↔ χ))
6	4, 5	bitri 240	. 2 ⊢ (((φ ↔ ψ) ↔ χ) ↔ ¬ ((φ ⊻ ψ) ↔ χ))
7	1, 2, 6	3bitr4i 268	1 ⊢ (hadd(φ, ψ, χ) ↔ ((φ ↔ ψ) ↔ χ))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 176   ⊻ wxo 1304  haddwhad 1378

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

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

This theorem is used by:  had1  1402