Proof of Theorem sotric
| Step | Hyp | Ref
| Expression |
| 1 | | breq2 2066 |
. . . . . . . 8
⊢ (B =
C → (BRB ↔ BRC)) |
| 2 | 1 | negbid 463 |
. . . . . . 7
⊢ (B =
C → (¬ BRB ↔ ¬ BRC)) |
| 3 | | sonr 2143 |
. . . . . . 7
⊢ ((R Or
A ∧ B ∈ A)
→ ¬ BRB) |
| 4 | 2, 3 | syl5bi 183 |
. . . . . 6
⊢ (B =
C → ((R Or A ∧
B ∈ A) → ¬ BRC)) |
| 5 | 4 | com12 13 |
. . . . 5
⊢ ((R Or
A ∧ B ∈ A)
→ (B = C → ¬ BRC)) |
| 6 | 5 | adantrr 312 |
. . . 4
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → (B =
C → ¬ BRC)) |
| 7 | | so2nr 2146 |
. . . . . 6
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → ¬ (BRC ∧ CRB)) |
| 8 | | imnan 207 |
. . . . . 6
⊢ ((BRC → ¬ CRB) ↔ ¬ (BRC ∧ CRB)) |
| 9 | 7, 8 | sylibr 175 |
. . . . 5
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → (BRC → ¬ CRB)) |
| 10 | 9 | con2d 83 |
. . . 4
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → (CRB → ¬ BRC)) |
| 11 | 6, 10 | jaod 329 |
. . 3
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → ((B =
C ∨ CRB) → ¬ BRC)) |
| 12 | | solin 2145 |
. . . 4
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → (BRC ∨ B =
C ∨ CRB)) |
| 13 | | 3orass 584 |
. . . . 5
⊢ ((BRC ∨ B =
C ∨ CRB) ↔ (BRC ∨ (B =
C ∨ CRB))) |
| 14 | | df-or 197 |
. . . . 5
⊢ ((BRC ∨ (B =
C ∨ CRB)) ↔ (¬ BRC → (B =
C ∨ CRB))) |
| 15 | 13, 14 | bitr 151 |
. . . 4
⊢ ((BRC ∨ B =
C ∨ CRB) ↔ (¬ BRC → (B =
C ∨ CRB))) |
| 16 | 12, 15 | sylib 173 |
. . 3
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → (¬ BRC → (B =
C ∨ CRB))) |
| 17 | 11, 16 | impbid 397 |
. 2
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → ((B =
C ∨ CRB) ↔ ¬ BRC)) |
| 18 | 17 | bicon2d 404 |
1
⊢ ((R Or
A ∧ (B ∈ A ∧
C ∈ A)) → (BRC ↔ ¬ (B = C ∨
CRB))) |