Metamath Proof Explorer
Description: Version of addcomli for natural numbers. (Contributed by Steven
Nguyen, 1-Aug-2023)
|
|
Ref |
Expression |
|
Hypotheses |
nnaddcomli.1 |
⊢ 𝐴 ∈ ℕ |
|
|
nnaddcomli.2 |
⊢ 𝐵 ∈ ℕ |
|
|
nnaddcomli.3 |
⊢ ( 𝐴 + 𝐵 ) = 𝐶 |
|
Assertion |
nnaddcomli |
⊢ ( 𝐵 + 𝐴 ) = 𝐶 |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
nnaddcomli.1 |
⊢ 𝐴 ∈ ℕ |
2 |
|
nnaddcomli.2 |
⊢ 𝐵 ∈ ℕ |
3 |
|
nnaddcomli.3 |
⊢ ( 𝐴 + 𝐵 ) = 𝐶 |
4 |
|
nnaddcom |
⊢ ( ( 𝐵 ∈ ℕ ∧ 𝐴 ∈ ℕ ) → ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐵 ) ) |
5 |
2 1 4
|
mp2an |
⊢ ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐵 ) |
6 |
5 3
|
eqtri |
⊢ ( 𝐵 + 𝐴 ) = 𝐶 |