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 ∈ ℂ
4 ax-1cn ⊢ 1 ∈ ℂ
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