Metamath Proof Explorer


Theorem 6p5lem

Description: Lemma for 6p5e11 and related theorems. (Contributed by Mario Carneiro, 19-Apr-2015)

Ref Expression
Hypotheses 6p5lem.1
|- A e. NN0
6p5lem.2
|- D e. NN0
6p5lem.3
|- E e. NN0
6p5lem.4
|- B = ( D + 1 )
6p5lem.5
|- C = ( E + 1 )
6p5lem.6
|- ( A + D ) = ; 1 E
Assertion 6p5lem
|- ( A + B ) = ; 1 C

Proof

Step Hyp Ref Expression
1 6p5lem.1
 |-  A e. NN0
2 6p5lem.2
 |-  D e. NN0
3 6p5lem.3
 |-  E e. NN0
4 6p5lem.4
 |-  B = ( D + 1 )
5 6p5lem.5
 |-  C = ( E + 1 )
6 6p5lem.6
 |-  ( A + D ) = ; 1 E
7 4 oveq2i
 |-  ( A + B ) = ( A + ( D + 1 ) )
8 1 nn0cni
 |-  A e. CC
9 2 nn0cni
 |-  D e. CC
10 ax-1cn
 |-  1 e. CC
11 8 9 10 addassi
 |-  ( ( A + D ) + 1 ) = ( A + ( D + 1 ) )
12 1nn0
 |-  1 e. NN0
13 5 eqcomi
 |-  ( E + 1 ) = C
14 12 3 13 6 decsuc
 |-  ( ( A + D ) + 1 ) = ; 1 C
15 7 11 14 3eqtr2i
 |-  ( A + B ) = ; 1 C