Theorem hadbi 1600

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 1504	. 2 ⊢ (((𝜑 ⊻ 𝜓) ⊻ 𝜒) ↔ ¬ ((𝜑 ⊻ 𝜓) ↔ 𝜒))
2		df-had 1596	. 2 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ⊻ 𝜓) ⊻ 𝜒))
3		xnor 1505	. . . 4 ⊢ ((𝜑 ↔ 𝜓) ↔ ¬ (𝜑 ⊻ 𝜓))
4	3	bibi1i 338	. . 3 ⊢ (((𝜑 ↔ 𝜓) ↔ 𝜒) ↔ (¬ (𝜑 ⊻ 𝜓) ↔ 𝜒))
5		nbbn 384	. . 3 ⊢ ((¬ (𝜑 ⊻ 𝜓) ↔ 𝜒) ↔ ¬ ((𝜑 ⊻ 𝜓) ↔ 𝜒))
6	4, 5	bitri 274	. 2 ⊢ (((𝜑 ↔ 𝜓) ↔ 𝜒) ↔ ¬ ((𝜑 ⊻ 𝜓) ↔ 𝜒))
7	1, 2, 6	3bitr4i 302	1 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ↔ 𝜓) ↔ 𝜒))

Syntax hints:  ¬ wn 3   ↔ wb 205   ⊻ wxo 1503  haddwhad 1595

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 206  df-xor 1504  df-had 1596

This theorem is referenced by:  hadcoma  1601  hadnot  1605  had1  1606