Step | Hyp | Ref
| Expression |
1 | | ssun1 3427 |
. . . . 5
⊢ Phi A ⊆ ( Phi A ∪ {0c}) |
2 | 1 | sseli 3270 |
. . . 4
⊢ (z ∈ Phi A →
z ∈
( Phi A ∪
{0c})) |
3 | | eleq2 2414 |
. . . 4
⊢ (( Phi A ∪
{0c}) = ( Phi B ∪ {0c}) → (z ∈ ( Phi A ∪
{0c}) ↔ z ∈ ( Phi B ∪ {0c}))) |
4 | 2, 3 | syl5ib 210 |
. . 3
⊢ (( Phi A ∪
{0c}) = ( Phi B ∪ {0c}) → (z ∈ Phi A →
z ∈
( Phi B ∪
{0c}))) |
5 | | 0cnelphi 4598 |
. . . . . 6
⊢ ¬
0c ∈
Phi A |
6 | | eleq1 2413 |
. . . . . 6
⊢ (z = 0c → (z ∈ Phi A ↔
0c ∈
Phi A)) |
7 | 5, 6 | mtbiri 294 |
. . . . 5
⊢ (z = 0c → ¬ z ∈ Phi A) |
8 | 7 | con2i 112 |
. . . 4
⊢ (z ∈ Phi A → ¬
z = 0c) |
9 | 8 | a1i 10 |
. . 3
⊢ (( Phi A ∪
{0c}) = ( Phi B ∪ {0c}) → (z ∈ Phi A → ¬
z = 0c)) |
10 | | elun 3221 |
. . . . . . 7
⊢ (z ∈ ( Phi B ∪
{0c}) ↔ (z ∈ Phi B ∨ z ∈
{0c})) |
11 | | df-sn 3742 |
. . . . . . . . 9
⊢
{0c} = {z ∣ z =
0c} |
12 | 11 | abeq2i 2461 |
. . . . . . . 8
⊢ (z ∈
{0c} ↔ z =
0c) |
13 | 12 | orbi2i 505 |
. . . . . . 7
⊢ ((z ∈ Phi B ∨ z ∈ {0c}) ↔ (z ∈ Phi B ∨ z =
0c)) |
14 | 10, 13 | bitri 240 |
. . . . . 6
⊢ (z ∈ ( Phi B ∪
{0c}) ↔ (z ∈ Phi B ∨ z = 0c)) |
15 | 14 | biimpi 186 |
. . . . 5
⊢ (z ∈ ( Phi B ∪
{0c}) → (z ∈ Phi B ∨ z = 0c)) |
16 | 15 | orcomd 377 |
. . . 4
⊢ (z ∈ ( Phi B ∪
{0c}) → (z =
0c ∨ z ∈ Phi B)) |
17 | 16 | ord 366 |
. . 3
⊢ (z ∈ ( Phi B ∪
{0c}) → (¬ z =
0c → z ∈ Phi B)) |
18 | 4, 9, 17 | ee22 1362 |
. 2
⊢ (( Phi A ∪
{0c}) = ( Phi B ∪ {0c}) → (z ∈ Phi A →
z ∈ Phi B)) |
19 | 18 | ssrdv 3279 |
1
⊢ (( Phi A ∪
{0c}) = ( Phi B ∪ {0c}) → Phi A ⊆ Phi B) |