Proof of Theorem preqr1g
Step | Hyp | Ref
| Expression |
1 | | prid1g 3680 |
. . . . . . 7
⊢ (𝐴 ∈ V → 𝐴 ∈ {𝐴, 𝐶}) |
2 | | eleq2 2230 |
. . . . . . 7
⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐴 ∈ {𝐴, 𝐶} ↔ 𝐴 ∈ {𝐵, 𝐶})) |
3 | 1, 2 | syl5ibcom 154 |
. . . . . 6
⊢ (𝐴 ∈ V → ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐴 ∈ {𝐵, 𝐶})) |
4 | | elprg 3596 |
. . . . . 6
⊢ (𝐴 ∈ V → (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
5 | 3, 4 | sylibd 148 |
. . . . 5
⊢ (𝐴 ∈ V → ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
6 | 5 | adantr 274 |
. . . 4
⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ V) → ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
7 | 6 | imp 123 |
. . 3
⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ {𝐴, 𝐶} = {𝐵, 𝐶}) → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶)) |
8 | | prid1g 3680 |
. . . . . . 7
⊢ (𝐵 ∈ V → 𝐵 ∈ {𝐵, 𝐶}) |
9 | | eleq2 2230 |
. . . . . . 7
⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 ∈ {𝐴, 𝐶} ↔ 𝐵 ∈ {𝐵, 𝐶})) |
10 | 8, 9 | syl5ibrcom 156 |
. . . . . 6
⊢ (𝐵 ∈ V → ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐵 ∈ {𝐴, 𝐶})) |
11 | | elprg 3596 |
. . . . . 6
⊢ (𝐵 ∈ V → (𝐵 ∈ {𝐴, 𝐶} ↔ (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))) |
12 | 10, 11 | sylibd 148 |
. . . . 5
⊢ (𝐵 ∈ V → ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))) |
13 | 12 | adantl 275 |
. . . 4
⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ V) → ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))) |
14 | 13 | imp 123 |
. . 3
⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ {𝐴, 𝐶} = {𝐵, 𝐶}) → (𝐵 = 𝐴 ∨ 𝐵 = 𝐶)) |
15 | | eqcom 2167 |
. . 3
⊢ (𝐴 = 𝐵 ↔ 𝐵 = 𝐴) |
16 | | eqeq2 2175 |
. . 3
⊢ (𝐴 = 𝐶 → (𝐵 = 𝐴 ↔ 𝐵 = 𝐶)) |
17 | 7, 14, 15, 16 | oplem1 965 |
. 2
⊢ (((𝐴 ∈ V ∧ 𝐵 ∈ V) ∧ {𝐴, 𝐶} = {𝐵, 𝐶}) → 𝐴 = 𝐵) |
18 | 17 | ex 114 |
1
⊢ ((𝐴 ∈ V ∧ 𝐵 ∈ V) → ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐴 = 𝐵)) |