Proof of Theorem ssprsseq
| Step | Hyp | Ref
| Expression |
| 1 | | ssprss 3800 |
. . . 4
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷)))) |
| 2 | 1 | 3adant3 1020 |
. . 3
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷)))) |
| 3 | | eqneqall 2387 |
. . . . . . . 8
⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐵 → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 4 | | eqtr3 2226 |
. . . . . . . 8
⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐶) → 𝐴 = 𝐵) |
| 5 | 3, 4 | syl11 31 |
. . . . . . 7
⊢ (𝐴 ≠ 𝐵 → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐶) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 6 | 5 | 3ad2ant3 1023 |
. . . . . 6
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐶) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 7 | 6 | com12 30 |
. . . . 5
⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐶) → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 8 | | preq12 3717 |
. . . . . . 7
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → {𝐴, 𝐵} = {𝐷, 𝐶}) |
| 9 | | prcom 3714 |
. . . . . . 7
⊢ {𝐷, 𝐶} = {𝐶, 𝐷} |
| 10 | 8, 9 | eqtrdi 2255 |
. . . . . 6
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → {𝐴, 𝐵} = {𝐶, 𝐷}) |
| 11 | 10 | a1d 22 |
. . . . 5
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 12 | | preq12 3717 |
. . . . . 6
⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝐴, 𝐵} = {𝐶, 𝐷}) |
| 13 | 12 | a1d 22 |
. . . . 5
⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 14 | | eqtr3 2226 |
. . . . . . . 8
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐷) → 𝐴 = 𝐵) |
| 15 | 3, 14 | syl11 31 |
. . . . . . 7
⊢ (𝐴 ≠ 𝐵 → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐷) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 16 | 15 | 3ad2ant3 1023 |
. . . . . 6
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐷) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 17 | 16 | com12 30 |
. . . . 5
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐷) → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 18 | 7, 11, 13, 17 | ccase 967 |
. . . 4
⊢ (((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷)) → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 19 | 18 | com12 30 |
. . 3
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → (((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) ∧ (𝐵 = 𝐶 ∨ 𝐵 = 𝐷)) → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 20 | 2, 19 | sylbid 150 |
. 2
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} → {𝐴, 𝐵} = {𝐶, 𝐷})) |
| 21 | | eqimss 3251 |
. 2
⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → {𝐴, 𝐵} ⊆ {𝐶, 𝐷}) |
| 22 | 20, 21 | impbid1 142 |
1
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) → ({𝐴, 𝐵} ⊆ {𝐶, 𝐷} ↔ {𝐴, 𝐵} = {𝐶, 𝐷})) |