Metamath Proof Explorer


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 ∈ ℂ
4 4cn 4 ∈ ℂ
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