Proof of Theorem sbccomlem
| Step | Hyp | Ref
 | Expression | 
| 1 |   | excom 1678 | 
. . . 4
⊢
(∃𝑥∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)) ↔ ∃𝑦∃𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑))) | 
| 2 |   | exdistr 1924 | 
. . . 4
⊢
(∃𝑥∃𝑦(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑))) | 
| 3 |   | an12 561 | 
. . . . . . 7
⊢ ((𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)) ↔ (𝑦 = 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑))) | 
| 4 | 3 | exbii 1619 | 
. . . . . 6
⊢
(∃𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)) ↔ ∃𝑥(𝑦 = 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑))) | 
| 5 |   | 19.42v 1921 | 
. . . . . 6
⊢
(∃𝑥(𝑦 = 𝐵 ∧ (𝑥 = 𝐴 ∧ 𝜑)) ↔ (𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))) | 
| 6 | 4, 5 | bitri 184 | 
. . . . 5
⊢
(∃𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)) ↔ (𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))) | 
| 7 | 6 | exbii 1619 | 
. . . 4
⊢
(∃𝑦∃𝑥(𝑥 = 𝐴 ∧ (𝑦 = 𝐵 ∧ 𝜑)) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))) | 
| 8 | 1, 2, 7 | 3bitr3i 210 | 
. . 3
⊢
(∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))) | 
| 9 |   | sbc5 3013 | 
. . 3
⊢
([𝐴 / 𝑥]∃𝑦(𝑦 = 𝐵 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑))) | 
| 10 |   | sbc5 3013 | 
. . 3
⊢
([𝐵 / 𝑦]∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑦(𝑦 = 𝐵 ∧ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))) | 
| 11 | 8, 9, 10 | 3bitr4i 212 | 
. 2
⊢
([𝐴 / 𝑥]∃𝑦(𝑦 = 𝐵 ∧ 𝜑) ↔ [𝐵 / 𝑦]∃𝑥(𝑥 = 𝐴 ∧ 𝜑)) | 
| 12 |   | sbc5 3013 | 
. . 3
⊢
([𝐵 / 𝑦]𝜑 ↔ ∃𝑦(𝑦 = 𝐵 ∧ 𝜑)) | 
| 13 | 12 | sbcbii 3049 | 
. 2
⊢
([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [𝐴 / 𝑥]∃𝑦(𝑦 = 𝐵 ∧ 𝜑)) | 
| 14 |   | sbc5 3013 | 
. . 3
⊢
([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)) | 
| 15 | 14 | sbcbii 3049 | 
. 2
⊢
([𝐵 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐵 / 𝑦]∃𝑥(𝑥 = 𝐴 ∧ 𝜑)) | 
| 16 | 11, 13, 15 | 3bitr4i 212 | 
1
⊢
([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [𝐵 / 𝑦][𝐴 / 𝑥]𝜑) |