Metamath Proof Explorer
		
		
		
		Description:  Commutation law for triple conjunction.  (Contributed by NM, 21-Apr-1994)
     (Proof shortened by Wolf Lammen, 5-Jun-2022)
		
			
				
					|  |  | Ref | Expression | 
				
					|  | Assertion | 3ancoma | ⊢  ( ( 𝜑  ∧  𝜓  ∧  𝜒 )  ↔  ( 𝜓  ∧  𝜑  ∧  𝜒 ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | 3anan12 | ⊢ ( ( 𝜑  ∧  𝜓  ∧  𝜒 )  ↔  ( 𝜓  ∧  ( 𝜑  ∧  𝜒 ) ) ) | 
						
							| 2 |  | 3anass | ⊢ ( ( 𝜓  ∧  𝜑  ∧  𝜒 )  ↔  ( 𝜓  ∧  ( 𝜑  ∧  𝜒 ) ) ) | 
						
							| 3 | 1 2 | bitr4i | ⊢ ( ( 𝜑  ∧  𝜓  ∧  𝜒 )  ↔  ( 𝜓  ∧  𝜑  ∧  𝜒 ) ) |