Proof of Theorem sspr
| Step | Hyp | Ref | Expression | 
|---|
| 1 |  | uncom 4158 | . . . . 5
⊢ (∅
∪ {𝐵, 𝐶}) = ({𝐵, 𝐶} ∪ ∅) | 
| 2 |  | un0 4394 | . . . . 5
⊢ ({𝐵, 𝐶} ∪ ∅) = {𝐵, 𝐶} | 
| 3 | 1, 2 | eqtri 2765 | . . . 4
⊢ (∅
∪ {𝐵, 𝐶}) = {𝐵, 𝐶} | 
| 4 | 3 | sseq2i 4013 | . . 3
⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 ⊆ {𝐵, 𝐶}) | 
| 5 |  | 0ss 4400 | . . . 4
⊢ ∅
⊆ 𝐴 | 
| 6 | 5 | biantrur 530 | . . 3
⊢ (𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}) ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}))) | 
| 7 | 4, 6 | bitr3i 277 | . 2
⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ (∅ ⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶}))) | 
| 8 |  | ssunpr 4834 | . 2
⊢ ((∅
⊆ 𝐴 ∧ 𝐴 ⊆ (∅ ∪ {𝐵, 𝐶})) ↔ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶})))) | 
| 9 |  | uncom 4158 | . . . . . 6
⊢ (∅
∪ {𝐵}) = ({𝐵} ∪
∅) | 
| 10 |  | un0 4394 | . . . . . 6
⊢ ({𝐵} ∪ ∅) = {𝐵} | 
| 11 | 9, 10 | eqtri 2765 | . . . . 5
⊢ (∅
∪ {𝐵}) = {𝐵} | 
| 12 | 11 | eqeq2i 2750 | . . . 4
⊢ (𝐴 = (∅ ∪ {𝐵}) ↔ 𝐴 = {𝐵}) | 
| 13 | 12 | orbi2i 913 | . . 3
⊢ ((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ↔ (𝐴 = ∅ ∨ 𝐴 = {𝐵})) | 
| 14 |  | uncom 4158 | . . . . . 6
⊢ (∅
∪ {𝐶}) = ({𝐶} ∪
∅) | 
| 15 |  | un0 4394 | . . . . . 6
⊢ ({𝐶} ∪ ∅) = {𝐶} | 
| 16 | 14, 15 | eqtri 2765 | . . . . 5
⊢ (∅
∪ {𝐶}) = {𝐶} | 
| 17 | 16 | eqeq2i 2750 | . . . 4
⊢ (𝐴 = (∅ ∪ {𝐶}) ↔ 𝐴 = {𝐶}) | 
| 18 | 3 | eqeq2i 2750 | . . . 4
⊢ (𝐴 = (∅ ∪ {𝐵, 𝐶}) ↔ 𝐴 = {𝐵, 𝐶}) | 
| 19 | 17, 18 | orbi12i 915 | . . 3
⊢ ((𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶})) ↔ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶})) | 
| 20 | 13, 19 | orbi12i 915 | . 2
⊢ (((𝐴 = ∅ ∨ 𝐴 = (∅ ∪ {𝐵})) ∨ (𝐴 = (∅ ∪ {𝐶}) ∨ 𝐴 = (∅ ∪ {𝐵, 𝐶}))) ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}))) | 
| 21 | 7, 8, 20 | 3bitri 297 | 1
⊢ (𝐴 ⊆ {𝐵, 𝐶} ↔ ((𝐴 = ∅ ∨ 𝐴 = {𝐵}) ∨ (𝐴 = {𝐶} ∨ 𝐴 = {𝐵, 𝐶}))) |