Metamath Proof Explorer


Theorem rspcsbnea

Description: Special case related to rspsbc . (Contributed by metakunt, 5-May-2025)

Ref Expression
Assertion rspcsbnea ⊢ A ∈ B ∧ ∀ x ∈ B C ≠ D → ⦋ A / x⦌ C ≠ D

Proof

Step Hyp Ref Expression
1 rspsbc ⊢ A ∈ B → ∀ x ∈ B C ≠ D → [˙A / x]˙ C ≠ D
2 df-ne ⊢ C ≠ D ↔ ¬ C = D
3 2 sbcbii ⊢ [˙A / x]˙ C ≠ D ↔ [˙A / x]˙ ¬ C = D
4 3 a1i ⊢ A ∈ B → [˙A / x]˙ C ≠ D ↔ [˙A / x]˙ ¬ C = D
5 sbcng ⊢ A ∈ B → [˙A / x]˙ ¬ C = D ↔ ¬ [˙A / x]˙ C = D
6 sbceq1g ⊢ A ∈ B → [˙A / x]˙ C = D ↔ ⦋ A / x⦌ C = D
7 6 notbid ⊢ A ∈ B → ¬ [˙A / x]˙ C = D ↔ ¬ ⦋ A / x⦌ C = D
8 5 7 bitrd ⊢ A ∈ B → [˙A / x]˙ ¬ C = D ↔ ¬ ⦋ A / x⦌ C = D
9 4 8 bitrd ⊢ A ∈ B → [˙A / x]˙ C ≠ D ↔ ¬ ⦋ A / x⦌ C = D
10 biidd ⊢ A ∈ B → ⦋ A / x⦌ C = D ↔ ⦋ A / x⦌ C = D
11 10 necon3bbid ⊢ A ∈ B → ¬ ⦋ A / x⦌ C = D ↔ ⦋ A / x⦌ C ≠ D
12 9 11 bitrd ⊢ A ∈ B → [˙A / x]˙ C ≠ D ↔ ⦋ A / x⦌ C ≠ D
13 1 12 sylibd ⊢ A ∈ B → ∀ x ∈ B C ≠ D → ⦋ A / x⦌ C ≠ D
14 13 imp ⊢ A ∈ B ∧ ∀ x ∈ B C ≠ D → ⦋ A / x⦌ C ≠ D