Proof of Theorem sbi2v
| Step | Hyp | Ref
 | Expression | 
| 1 |   | 19.38 1690 | 
. . 3
⊢
((∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓)) → ∀𝑥((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜓))) | 
| 2 |   | pm3.3 261 | 
. . . . 5
⊢ (((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜓)) → (𝑥 = 𝑦 → (𝜑 → (𝑥 = 𝑦 → 𝜓)))) | 
| 3 |   | pm2.04 82 | 
. . . . 5
⊢ ((𝜑 → (𝑥 = 𝑦 → 𝜓)) → (𝑥 = 𝑦 → (𝜑 → 𝜓))) | 
| 4 | 2, 3 | syli 37 | 
. . . 4
⊢ (((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜓)) → (𝑥 = 𝑦 → (𝜑 → 𝜓))) | 
| 5 | 4 | alimi 1469 | 
. . 3
⊢
(∀𝑥((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜓)) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓))) | 
| 6 | 1, 5 | syl 14 | 
. 2
⊢
((∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓)) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓))) | 
| 7 |   | sb5 1902 | 
. . 3
⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) | 
| 8 |   | sb6 1901 | 
. . 3
⊢ ([𝑦 / 𝑥]𝜓 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓)) | 
| 9 | 7, 8 | imbi12i 239 | 
. 2
⊢ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ↔ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓))) | 
| 10 |   | sb6 1901 | 
. 2
⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓))) | 
| 11 | 6, 9, 10 | 3imtr4i 201 | 
1
⊢ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) → [𝑦 / 𝑥](𝜑 → 𝜓)) |