Metamath Proof Explorer


Theorem 1p2e3

Description: 1 + 2 = 3. For a shorter proof using addcomli , see 1p2e3ALT . (Contributed by David A. Wheeler, 8-Dec-2018) Reduce dependencies on axioms. (Revised by Steven Nguyen, 12-Dec-2022)

Ref Expression
Assertion 1p2e3
|- ( 1 + 2 ) = 3

Proof

Step Hyp Ref Expression
1 df-2
 |-  2 = ( 1 + 1 )
2 1 oveq2i
 |-  ( 1 + 2 ) = ( 1 + ( 1 + 1 ) )
3 ax-1cn
 |-  1 e. CC
4 3 3 3 addassi
 |-  ( ( 1 + 1 ) + 1 ) = ( 1 + ( 1 + 1 ) )
5 1p1e2
 |-  ( 1 + 1 ) = 2
6 5 oveq1i
 |-  ( ( 1 + 1 ) + 1 ) = ( 2 + 1 )
7 2p1e3
 |-  ( 2 + 1 ) = 3
8 6 7 eqtri
 |-  ( ( 1 + 1 ) + 1 ) = 3
9 2 4 8 3eqtr2i
 |-  ( 1 + 2 ) = 3