Metamath Proof Explorer


Theorem dpadd

Description: Addition with one decimal. (Contributed by Thierry Arnoux, 27-Dec-2021)

Ref Expression
Hypotheses dpmul.a 𝐴 ∈ ℕ0
dpmul.b 𝐵 ∈ ℕ0
dpmul.c 𝐶 ∈ ℕ0
dpmul.d 𝐷 ∈ ℕ0
dpmul.e 𝐸 ∈ ℕ0
dpadd.f 𝐹 ∈ ℕ0
dpadd.1 ( 𝐴 𝐵 + 𝐶 𝐷 ) = 𝐸 𝐹
Assertion dpadd ( ( 𝐴 . 𝐵 ) + ( 𝐶 . 𝐷 ) ) = ( 𝐸 . 𝐹 )

Proof

Step Hyp Ref Expression
1 dpmul.a 𝐴 ∈ ℕ0
2 dpmul.b 𝐵 ∈ ℕ0
3 dpmul.c 𝐶 ∈ ℕ0
4 dpmul.d 𝐷 ∈ ℕ0
5 dpmul.e 𝐸 ∈ ℕ0
6 dpadd.f 𝐹 ∈ ℕ0
7 dpadd.1 ( 𝐴 𝐵 + 𝐶 𝐷 ) = 𝐸 𝐹
8 1 2 deccl 𝐴 𝐵 ∈ ℕ0
9 8 nn0cni 𝐴 𝐵 ∈ ℂ
10 3 4 deccl 𝐶 𝐷 ∈ ℕ0
11 10 nn0cni 𝐶 𝐷 ∈ ℂ
12 10nn 1 0 ∈ ℕ
13 12 nncni 1 0 ∈ ℂ
14 12 nnne0i 1 0 ≠ 0
15 9 11 13 14 divdiri ( ( 𝐴 𝐵 + 𝐶 𝐷 ) / 1 0 ) = ( ( 𝐴 𝐵 / 1 0 ) + ( 𝐶 𝐷 / 1 0 ) )
16 7 oveq1i ( ( 𝐴 𝐵 + 𝐶 𝐷 ) / 1 0 ) = ( 𝐸 𝐹 / 1 0 )
17 15 16 eqtr3i ( ( 𝐴 𝐵 / 1 0 ) + ( 𝐶 𝐷 / 1 0 ) ) = ( 𝐸 𝐹 / 1 0 )
18 2 nn0rei 𝐵 ∈ ℝ
19 1 18 decdiv10 ( 𝐴 𝐵 / 1 0 ) = ( 𝐴 . 𝐵 )
20 4 nn0rei 𝐷 ∈ ℝ
21 3 20 decdiv10 ( 𝐶 𝐷 / 1 0 ) = ( 𝐶 . 𝐷 )
22 19 21 oveq12i ( ( 𝐴 𝐵 / 1 0 ) + ( 𝐶 𝐷 / 1 0 ) ) = ( ( 𝐴 . 𝐵 ) + ( 𝐶 . 𝐷 ) )
23 6 nn0rei 𝐹 ∈ ℝ
24 5 23 decdiv10 ( 𝐸 𝐹 / 1 0 ) = ( 𝐸 . 𝐹 )
25 17 22 24 3eqtr3i ( ( 𝐴 . 𝐵 ) + ( 𝐶 . 𝐷 ) ) = ( 𝐸 . 𝐹 )