Proof of Theorem add4
Step | Hyp | Ref
| Expression |
1 | | add12 8077 |
. . . . 5
⊢ ((𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐵 + (𝐶 + 𝐷)) = (𝐶 + (𝐵 + 𝐷))) |
2 | 1 | 3expb 1199 |
. . . 4
⊢ ((𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐵 + (𝐶 + 𝐷)) = (𝐶 + (𝐵 + 𝐷))) |
3 | 2 | oveq2d 5869 |
. . 3
⊢ ((𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐴 + (𝐵 + (𝐶 + 𝐷))) = (𝐴 + (𝐶 + (𝐵 + 𝐷)))) |
4 | 3 | adantll 473 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐴 + (𝐵 + (𝐶 + 𝐷))) = (𝐴 + (𝐶 + (𝐵 + 𝐷)))) |
5 | | addcl 7899 |
. . 3
⊢ ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 + 𝐷) ∈ ℂ) |
6 | | addass 7904 |
. . . 4
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 + 𝐷) ∈ ℂ) → ((𝐴 + 𝐵) + (𝐶 + 𝐷)) = (𝐴 + (𝐵 + (𝐶 + 𝐷)))) |
7 | 6 | 3expa 1198 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 + 𝐷) ∈ ℂ) → ((𝐴 + 𝐵) + (𝐶 + 𝐷)) = (𝐴 + (𝐵 + (𝐶 + 𝐷)))) |
8 | 5, 7 | sylan2 284 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 + 𝐵) + (𝐶 + 𝐷)) = (𝐴 + (𝐵 + (𝐶 + 𝐷)))) |
9 | | addcl 7899 |
. . . 4
⊢ ((𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐵 + 𝐷) ∈ ℂ) |
10 | | addass 7904 |
. . . . 5
⊢ ((𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ (𝐵 + 𝐷) ∈ ℂ) → ((𝐴 + 𝐶) + (𝐵 + 𝐷)) = (𝐴 + (𝐶 + (𝐵 + 𝐷)))) |
11 | 10 | 3expa 1198 |
. . . 4
⊢ (((𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐵 + 𝐷) ∈ ℂ) → ((𝐴 + 𝐶) + (𝐵 + 𝐷)) = (𝐴 + (𝐶 + (𝐵 + 𝐷)))) |
12 | 9, 11 | sylan2 284 |
. . 3
⊢ (((𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 + 𝐶) + (𝐵 + 𝐷)) = (𝐴 + (𝐶 + (𝐵 + 𝐷)))) |
13 | 12 | an4s 583 |
. 2
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 + 𝐶) + (𝐵 + 𝐷)) = (𝐴 + (𝐶 + (𝐵 + 𝐷)))) |
14 | 4, 8, 13 | 3eqtr4d 2213 |
1
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 + 𝐵) + (𝐶 + 𝐷)) = ((𝐴 + 𝐶) + (𝐵 + 𝐷))) |