Theorem hadbi 1528

Description: The adder sum 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 1457	. 2 ⊢ (((𝜑 ⊻ 𝜓) ⊻ 𝜒) ↔ ¬ ((𝜑 ⊻ 𝜓) ↔ 𝜒))
2		df-had 1524	. 2 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ⊻ 𝜓) ⊻ 𝜒))
3		xnor 1458	. . . 4 ⊢ ((𝜑 ↔ 𝜓) ↔ ¬ (𝜑 ⊻ 𝜓))
4	3	bibi1i 327	. . 3 ⊢ (((𝜑 ↔ 𝜓) ↔ 𝜒) ↔ (¬ (𝜑 ⊻ 𝜓) ↔ 𝜒))
5		nbbn 372	. . 3 ⊢ ((¬ (𝜑 ⊻ 𝜓) ↔ 𝜒) ↔ ¬ ((𝜑 ⊻ 𝜓) ↔ 𝜒))
6	4, 5	bitri 263	. 2 ⊢ (((𝜑 ↔ 𝜓) ↔ 𝜒) ↔ ¬ ((𝜑 ⊻ 𝜓) ↔ 𝜒))
7	1, 2, 6	3bitr4i 291	1 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ↔ 𝜓) ↔ 𝜒))

Syntax hints:  ¬ wn 3   ↔ wb 195   ⊻ wxo 1456  haddwhad 1523

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 196  df-xor 1457  df-had 1524

This theorem is referenced by:  hadnot  1532  had1  1533