Proof of Theorem example-E6.01a
| Step | Hyp | Ref
| Expression |
| 1 | | example-E6.01a.1 |
. 2
⊢ (𝑏 = 𝑏₁
→ (𝜓 ↔ 𝜓₁)) |
| 2 | | nfrv-P6 686 |
. . 3
⊢ Ⅎ𝑏∀𝑦(𝑦 ∈ 𝑥 →
𝑦 ∈ 𝑎) |
| 3 | | example-E6.01a.3 |
. . . . 5
⊢ (𝜓 ↔ ∀𝑦(𝑦 ∈ 𝑥 → 𝑦 ∈
𝑎)) |
| 4 | 3 | bisym-P3.33b.RC 299 |
. . . 4
⊢ (∀𝑦(𝑦 ∈ 𝑥 → 𝑦 ∈
𝑎) ↔ 𝜓) |
| 5 | 4 | nfrleq-P6 687 |
. . 3
⊢ (Ⅎ𝑏∀𝑦(𝑦 ∈ 𝑥 →
𝑦 ∈ 𝑎)
↔ Ⅎ𝑏𝜓) |
| 6 | 2, 5 | bimpf-P4.RC 532 |
. 2
⊢ Ⅎ𝑏𝜓 |
| 7 | | subelofl-P5.CL 639 |
. . . . . . 7
⊢ (𝑥 = 𝑦 →
(𝑥 ∈ 𝑏
↔ 𝑦 ∈ 𝑏)) |
| 8 | | subelofl-P5.CL 639 |
. . . . . . 7
⊢ (𝑥 = 𝑦 →
(𝑥 ∈ 𝑎
↔ 𝑦 ∈ 𝑎)) |
| 9 | 7, 8 | subimd-P3.40c 329 |
. . . . . 6
⊢ (𝑥 = 𝑦 →
((𝑥 ∈ 𝑏 → 𝑥 ∈
𝑎) ↔ (𝑦 ∈ 𝑏 →
𝑦 ∈ 𝑎))) |
| 10 | 9 | cbvallv-P5 659 |
. . . . 5
⊢ (∀𝑥(𝑥 ∈ 𝑏 → 𝑥 ∈
𝑎) ↔ ∀𝑦(𝑦 ∈ 𝑏 → 𝑦 ∈
𝑎)) |
| 11 | 10 | rcp-NDIMP0addall 207 |
. . . 4
⊢ (𝑏 = 𝑥 →
(∀𝑥(𝑥 ∈ 𝑏 →
𝑥 ∈ 𝑎)
↔ ∀𝑦(𝑦 ∈ 𝑏 →
𝑦 ∈ 𝑎))) |
| 12 | | subelofr-P5.CL 641 |
. . . . . 6
⊢ (𝑏 = 𝑥 →
(𝑦 ∈ 𝑏
↔ 𝑦 ∈ 𝑥)) |
| 13 | 12 | subiml-P3.40a 325 |
. . . . 5
⊢ (𝑏 = 𝑥 →
((𝑦 ∈ 𝑏 → 𝑦 ∈
𝑎) ↔ (𝑦 ∈ 𝑥 →
𝑦 ∈ 𝑎))) |
| 14 | 13 | suballv-P5 623 |
. . . 4
⊢ (𝑏 = 𝑥 →
(∀𝑦(𝑦 ∈ 𝑏 →
𝑦 ∈ 𝑎)
↔ ∀𝑦(𝑦 ∈ 𝑥 →
𝑦 ∈ 𝑎))) |
| 15 | 11, 14 | bitrns-P3.33c 302 |
. . 3
⊢ (𝑏 = 𝑥 →
(∀𝑥(𝑥 ∈ 𝑏 →
𝑥 ∈ 𝑎)
↔ ∀𝑦(𝑦 ∈ 𝑥 →
𝑦 ∈ 𝑎))) |
| 16 | | example-E6.01a.2 |
. . . . 5
⊢ (𝜑 ↔ ∀𝑥(𝑥 ∈ 𝑏 → 𝑥 ∈
𝑎)) |
| 17 | 16 | bisym-P3.33b.RC 299 |
. . . 4
⊢ (∀𝑥(𝑥 ∈ 𝑏 → 𝑥 ∈
𝑎) ↔ 𝜑) |
| 18 | 17, 4 | subbid-P3.41c.RC 337 |
. . 3
⊢ ((∀𝑥(𝑥 ∈ 𝑏 → 𝑥 ∈
𝑎) ↔ ∀𝑦(𝑦 ∈ 𝑥 → 𝑦 ∈
𝑎)) ↔ (𝜑
↔ 𝜓)) |
| 19 | 15, 18 | subimr2-P4.RC 543 |
. 2
⊢ (𝑏 = 𝑥 →
(𝜑 ↔ 𝜓)) |
| 20 | 1, 6, 19 | solvesub-P6a 704 |
1
⊢ (𝜓 ↔ ∀𝑏(𝑏 = 𝑥 → 𝜑)) |