Theorem hadcoma 1601

Description: Commutative law for the adder sum. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 17-Dec-2023.)

Assertion

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

Proof of Theorem hadcoma

Step	Hyp	Ref	Expression
1		bicom 221	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ (𝜓 ↔ 𝜑))
2	1	bibi1i 338	. 2 ⊢ (((𝜑 ↔ 𝜓) ↔ 𝜒) ↔ ((𝜓 ↔ 𝜑) ↔ 𝜒))
3		hadbi 1600	. 2 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜑 ↔ 𝜓) ↔ 𝜒))
4		hadbi 1600	. 2 ⊢ (hadd(𝜓, 𝜑, 𝜒) ↔ ((𝜓 ↔ 𝜑) ↔ 𝜒))
5	2, 3, 4	3bitr4i 302	1 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ hadd(𝜓, 𝜑, 𝜒))

Syntax hints:   ↔ wb 205  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:  hadrot  1604  sadcom  16098