Metamath Proof Explorer


Theorem 6p2e8

Description: 6 + 2 = 8. (Contributed by NM, 11-May-2004)

Ref Expression
Assertion 6p2e8
|- ( 6 + 2 ) = 8

Proof

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