Theorem 4p2e6

Description: 4 + 2 = 6. (Contributed by NM, 11-May-2004)

	Ref	Expression
Assertion	4p2e6	|- ( 4 + 2 ) = 6

Proof

Step	Hyp	Ref	Expression
1		df-2	|- 2 = ( 1 + 1 )
2	1	oveq2i	|- ( 4 + 2 ) = ( 4 + ( 1 + 1 ) )
3		4cn	|- 4 e. CC
4		ax-1cn	|- 1 e. CC
5	3 4 4	addassi	|- ( ( 4 + 1 ) + 1 ) = ( 4 + ( 1 + 1 ) )
6	2 5	eqtr4i	|- ( 4 + 2 ) = ( ( 4 + 1 ) + 1 )
7		df-5	|- 5 = ( 4 + 1 )
8	7	oveq1i	|- ( 5 + 1 ) = ( ( 4 + 1 ) + 1 )
9	6 8	eqtr4i	|- ( 4 + 2 ) = ( 5 + 1 )
10		df-6	|- 6 = ( 5 + 1 )
11	9 10	eqtr4i	|- ( 4 + 2 ) = 6