| Step |
Hyp |
Ref |
Expression |
| 1 |
|
chjcom |
⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ) → ( 𝐴 ∨ℋ 𝐵 ) = ( 𝐵 ∨ℋ 𝐴 ) ) |
| 2 |
1
|
3adant3 |
⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ∧ 𝐶 ∈ Cℋ ) → ( 𝐴 ∨ℋ 𝐵 ) = ( 𝐵 ∨ℋ 𝐴 ) ) |
| 3 |
2
|
oveq1d |
⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ∧ 𝐶 ∈ Cℋ ) → ( ( 𝐴 ∨ℋ 𝐵 ) ∨ℋ 𝐶 ) = ( ( 𝐵 ∨ℋ 𝐴 ) ∨ℋ 𝐶 ) ) |
| 4 |
|
chjass |
⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ∧ 𝐶 ∈ Cℋ ) → ( ( 𝐴 ∨ℋ 𝐵 ) ∨ℋ 𝐶 ) = ( 𝐴 ∨ℋ ( 𝐵 ∨ℋ 𝐶 ) ) ) |
| 5 |
|
chjass |
⊢ ( ( 𝐵 ∈ Cℋ ∧ 𝐴 ∈ Cℋ ∧ 𝐶 ∈ Cℋ ) → ( ( 𝐵 ∨ℋ 𝐴 ) ∨ℋ 𝐶 ) = ( 𝐵 ∨ℋ ( 𝐴 ∨ℋ 𝐶 ) ) ) |
| 6 |
5
|
3com12 |
⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ∧ 𝐶 ∈ Cℋ ) → ( ( 𝐵 ∨ℋ 𝐴 ) ∨ℋ 𝐶 ) = ( 𝐵 ∨ℋ ( 𝐴 ∨ℋ 𝐶 ) ) ) |
| 7 |
3 4 6
|
3eqtr3d |
⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ∧ 𝐶 ∈ Cℋ ) → ( 𝐴 ∨ℋ ( 𝐵 ∨ℋ 𝐶 ) ) = ( 𝐵 ∨ℋ ( 𝐴 ∨ℋ 𝐶 ) ) ) |