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Mirrors > Home > ILE Home > Th. List > eqrdav | GIF version |
Description: Deduce equality of classes from an equivalence of membership that depends on the membership variable. (Contributed by NM, 7-Nov-2008.) |
Ref | Expression |
---|---|
eqrdav.1 | ⊢ ((φ ∧ x ∈ A) → x ∈ 𝐶) |
eqrdav.2 | ⊢ ((φ ∧ x ∈ B) → x ∈ 𝐶) |
eqrdav.3 | ⊢ ((φ ∧ x ∈ 𝐶) → (x ∈ A ↔ x ∈ B)) |
Ref | Expression |
---|---|
eqrdav | ⊢ (φ → A = B) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqrdav.1 | . . . 4 ⊢ ((φ ∧ x ∈ A) → x ∈ 𝐶) | |
2 | eqrdav.3 | . . . . . 6 ⊢ ((φ ∧ x ∈ 𝐶) → (x ∈ A ↔ x ∈ B)) | |
3 | 2 | biimpd 132 | . . . . 5 ⊢ ((φ ∧ x ∈ 𝐶) → (x ∈ A → x ∈ B)) |
4 | 3 | impancom 247 | . . . 4 ⊢ ((φ ∧ x ∈ A) → (x ∈ 𝐶 → x ∈ B)) |
5 | 1, 4 | mpd 13 | . . 3 ⊢ ((φ ∧ x ∈ A) → x ∈ B) |
6 | eqrdav.2 | . . . 4 ⊢ ((φ ∧ x ∈ B) → x ∈ 𝐶) | |
7 | 2 | exbiri 364 | . . . . . 6 ⊢ (φ → (x ∈ 𝐶 → (x ∈ B → x ∈ A))) |
8 | 7 | com23 72 | . . . . 5 ⊢ (φ → (x ∈ B → (x ∈ 𝐶 → x ∈ A))) |
9 | 8 | imp 115 | . . . 4 ⊢ ((φ ∧ x ∈ B) → (x ∈ 𝐶 → x ∈ A)) |
10 | 6, 9 | mpd 13 | . . 3 ⊢ ((φ ∧ x ∈ B) → x ∈ A) |
11 | 5, 10 | impbida 528 | . 2 ⊢ (φ → (x ∈ A ↔ x ∈ B)) |
12 | 11 | eqrdv 2035 | 1 ⊢ (φ → A = B) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 97 ↔ wb 98 = wceq 1242 ∈ wcel 1390 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1333 ax-gen 1335 ax-17 1416 ax-ext 2019 |
This theorem depends on definitions: df-bi 110 df-cleq 2030 |
This theorem is referenced by: fzdifsuc 8713 |
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