Metamath Proof Explorer


Theorem 3p2e5

Description: 3 + 2 = 5. (Contributed by NM, 11-May-2004)

Ref Expression
Assertion 3p2e5
|- ( 3 + 2 ) = 5

Proof

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