Proof of Theorem fri
| Step | Hyp | Ref
| Expression |
| 1 | | df-fr 2169 |
. . 3
⊢ (R Fr
A ↔ ∀z((z ⊆
A ∧ ¬ z = ∅) → ∃x ∈ z
∀y ∈ z ¬ yRx)) |
| 2 | | fri.1 |
. . . 4
⊢ B
∈ V |
| 3 | | sseq1 1521 |
. . . . . 6
⊢ (z =
B → (z ⊆ A
↔ B ⊆ A)) |
| 4 | | cleq1 1107 |
. . . . . . 7
⊢ (z =
B → (z = ∅ ↔ B = ∅)) |
| 5 | 4 | negbid 463 |
. . . . . 6
⊢ (z =
B → (¬ z = ∅ ↔ ¬ B = ∅)) |
| 6 | 3, 5 | anbi12d 476 |
. . . . 5
⊢ (z =
B → ((z ⊆ A
∧ ¬ z = ∅) ↔ (B ⊆ A
∧ ¬ B = ∅))) |
| 7 | | raleq 1324 |
. . . . . 6
⊢ (z =
B → (∀y ∈ z ¬
yRx ↔
∀y ∈ B ¬ yRx)) |
| 8 | 7 | rexeqd 1328 |
. . . . 5
⊢ (z =
B → (∃x ∈ z
∀y ∈ z ¬ yRx ↔ ∃x ∈ B
∀y ∈ B ¬ yRx)) |
| 9 | 6, 8 | imbi12d 474 |
. . . 4
⊢ (z =
B → (((z ⊆ A
∧ ¬ z = ∅) →
∃x ∈ z ∀y
∈ z ¬ yRx) ↔ ((B
⊆ A ∧ ¬ B = ∅) → ∃x ∈ B
∀y ∈ B ¬ yRx))) |
| 10 | 2, 9 | cla4v 1400 |
. . 3
⊢ (∀z((z ⊆
A ∧ ¬ z = ∅) → ∃x ∈ z
∀y ∈ z ¬ yRx) → ((B
⊆ A ∧ ¬ B = ∅) → ∃x ∈ B
∀y ∈ B ¬ yRx)) |
| 11 | 1, 10 | sylbi 174 |
. 2
⊢ (R Fr
A → ((B ⊆ A
∧ ¬ B = ∅) →
∃x ∈ B ∀y
∈ B ¬ yRx)) |
| 12 | 11 | imp 277 |
1
⊢ ((R Fr
A ∧ (B ⊆ A
∧ ¬ B = ∅)) →
∃x ∈ B ∀y
∈ B ¬ yRx) |