Metamath Proof Explorer


Theorem 6p3e9

Description: 6 + 3 = 9. (Contributed by NM, 11-May-2004)

Ref Expression
Assertion 6p3e9
|- ( 6 + 3 ) = 9

Proof

Step Hyp Ref Expression
1 df-3
 |-  3 = ( 2 + 1 )
2 1 oveq2i
 |-  ( 6 + 3 ) = ( 6 + ( 2 + 1 ) )
3 6cn
 |-  6 e. CC
4 2cn
 |-  2 e. CC
5 ax-1cn
 |-  1 e. CC
6 3 4 5 addassi
 |-  ( ( 6 + 2 ) + 1 ) = ( 6 + ( 2 + 1 ) )
7 2 6 eqtr4i
 |-  ( 6 + 3 ) = ( ( 6 + 2 ) + 1 )
8 df-9
 |-  9 = ( 8 + 1 )
9 6p2e8
 |-  ( 6 + 2 ) = 8
10 9 oveq1i
 |-  ( ( 6 + 2 ) + 1 ) = ( 8 + 1 )
11 8 10 eqtr4i
 |-  9 = ( ( 6 + 2 ) + 1 )
12 7 11 eqtr4i
 |-  ( 6 + 3 ) = 9