Theorem 1p5e6

Description: 1 + 5 = 6. (Contributed by SN, 24-Aug-2026)

	Ref	Expression
Assertion	1p5e6	|- ( 1 + 5 ) = 6

Proof

Step	Hyp	Ref	Expression
1		df-5	|- 5 = ( 4 + 1 )
2	1	oveq2i	|- ( 1 + 5 ) = ( 1 + ( 4 + 1 ) )
3		ax-1cn	|- 1 e. CC
4		4cn	|- 4 e. CC
5	3 4 3	addassi	|- ( ( 1 + 4 ) + 1 ) = ( 1 + ( 4 + 1 ) )
6		1p4e5	|- ( 1 + 4 ) = 5
7	6	oveq1i	|- ( ( 1 + 4 ) + 1 ) = ( 5 + 1 )
8		5p1e6	|- ( 5 + 1 ) = 6
9	7 8	eqtri	|- ( ( 1 + 4 ) + 1 ) = 6
10	2 5 9	3eqtr2i	|- ( 1 + 5 ) = 6