Metamath Proof Explorer


Theorem 6p6e12

Description: 6 + 6 = 12. (Contributed by Mario Carneiro, 19-Apr-2015)

Ref Expression
Assertion 6p6e12
|- ( 6 + 6 ) = ; 1 2

Proof

Step Hyp Ref Expression
1 6nn0
 |-  6 e. NN0
2 5nn0
 |-  5 e. NN0
3 1nn0
 |-  1 e. NN0
4 df-6
 |-  6 = ( 5 + 1 )
5 df-2
 |-  2 = ( 1 + 1 )
6 6p5e11
 |-  ( 6 + 5 ) = ; 1 1
7 1 2 3 4 5 6 6p5lem
 |-  ( 6 + 6 ) = ; 1 2