Proof of Theorem phiall
Step | Hyp | Ref
| Expression |
1 | | neldifsn 3842 |
. . . . 5
⊢ ¬
0c ∈ (A ∖
{0c}) |
2 | | phiall.1 |
. . . . . . 7
⊢ A ∈
V |
3 | | snex 4112 |
. . . . . . 7
⊢
{0c} ∈
V |
4 | 2, 3 | difex 4108 |
. . . . . 6
⊢ (A ∖
{0c}) ∈ V |
5 | 4 | phialllem2 4618 |
. . . . 5
⊢ (¬
0c ∈ (A ∖
{0c}) → ∃x(A ∖ {0c}) =
Phi x) |
6 | 1, 5 | ax-mp 5 |
. . . 4
⊢ ∃x(A ∖
{0c}) = Phi x |
7 | | disjsn 3787 |
. . . . . . . . . 10
⊢ (((A ∖
{0c}) ∩ {0c}) = ∅ ↔ ¬ 0c ∈ (A ∖ {0c})) |
8 | 1, 7 | mpbir 200 |
. . . . . . . . 9
⊢ ((A ∖
{0c}) ∩ {0c}) = ∅ |
9 | | 0cnelphi 4598 |
. . . . . . . . . 10
⊢ ¬
0c ∈
Phi x |
10 | | disjsn 3787 |
. . . . . . . . . 10
⊢ (( Phi x ∩
{0c}) = ∅ ↔ ¬
0c ∈
Phi x) |
11 | 9, 10 | mpbir 200 |
. . . . . . . . 9
⊢ ( Phi x ∩
{0c}) = ∅ |
12 | 8, 11 | eqtr4i 2376 |
. . . . . . . 8
⊢ ((A ∖
{0c}) ∩ {0c}) = (
Phi x ∩
{0c}) |
13 | 12 | biantru 491 |
. . . . . . 7
⊢ (((A ∖
{0c}) ∪ {0c}) = (
Phi x ∪ {0c})
↔ (((A ∖ {0c}) ∪
{0c}) = ( Phi x ∪ {0c}) ∧ ((A ∖ {0c}) ∩
{0c}) = ( Phi x ∩ {0c}))) |
14 | | unineq 3506 |
. . . . . . 7
⊢ ((((A ∖
{0c}) ∪ {0c}) = (
Phi x ∪ {0c})
∧ ((A
∖ {0c}) ∩
{0c}) = ( Phi x ∩ {0c})) ↔ (A ∖
{0c}) = Phi x) |
15 | 13, 14 | bitri 240 |
. . . . . 6
⊢ (((A ∖
{0c}) ∪ {0c}) = (
Phi x ∪ {0c})
↔ (A ∖ {0c}) =
Phi x) |
16 | | difsnid 3855 |
. . . . . . 7
⊢
(0c ∈ A → ((A
∖ {0c}) ∪
{0c}) = A) |
17 | 16 | eqeq1d 2361 |
. . . . . 6
⊢
(0c ∈ A → (((A
∖ {0c}) ∪
{0c}) = ( Phi x ∪ {0c}) ↔ A = ( Phi x ∪ {0c}))) |
18 | 15, 17 | syl5bbr 250 |
. . . . 5
⊢
(0c ∈ A → ((A
∖ {0c}) = Phi x ↔
A = ( Phi
x ∪
{0c}))) |
19 | 18 | exbidv 1626 |
. . . 4
⊢
(0c ∈ A → (∃x(A ∖
{0c}) = Phi x ↔ ∃x A = ( Phi x ∪ {0c}))) |
20 | 6, 19 | mpbii 202 |
. . 3
⊢
(0c ∈ A → ∃x A = ( Phi x ∪ {0c})) |
21 | | olc 373 |
. . . 4
⊢ (A = ( Phi x ∪ {0c}) → (A = Phi x ∨ A = ( Phi x ∪ {0c}))) |
22 | 21 | eximi 1576 |
. . 3
⊢ (∃x A = ( Phi x ∪ {0c}) → ∃x(A = Phi x ∨ A = ( Phi x ∪ {0c}))) |
23 | 20, 22 | syl 15 |
. 2
⊢
(0c ∈ A → ∃x(A = Phi x ∨ A = ( Phi x ∪ {0c}))) |
24 | 2 | phialllem2 4618 |
. . 3
⊢ (¬
0c ∈ A → ∃x A = Phi x) |
25 | | orc 374 |
. . . 4
⊢ (A = Phi x → (A =
Phi x ∨ A = ( Phi x ∪
{0c}))) |
26 | 25 | eximi 1576 |
. . 3
⊢ (∃x A = Phi x → ∃x(A = Phi x ∨ A = ( Phi x ∪ {0c}))) |
27 | 24, 26 | syl 15 |
. 2
⊢ (¬
0c ∈ A → ∃x(A = Phi x ∨ A = ( Phi x ∪ {0c}))) |
28 | 23, 27 | pm2.61i 156 |
1
⊢ ∃x(A = Phi x ∨ A = ( Phi x ∪ {0c})) |