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| Mirrors > Home > ILE Home > Th. List > axext4 | GIF version | ||
| Description: A bidirectional version of Extensionality. Although this theorem "looks" like it is just a definition of equality, it requires the Axiom of Extensionality for its proof under our axiomatization. See the comments for ax-ext 2178. (Contributed by NM, 14-Nov-2008.) |
| Ref | Expression |
|---|---|
| axext4 | ⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | elequ2 2172 | . . 3 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) | |
| 2 | 1 | alrimiv 1888 | . 2 ⊢ (𝑥 = 𝑦 → ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) |
| 3 | axext3 2179 | . 2 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦) | |
| 4 | 2, 3 | impbii 126 | 1 ⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) |
| Colors of variables: wff set class |
| Syntax hints: ↔ wb 105 ∀wal 1362 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-5 1461 ax-gen 1463 ax-ie1 1507 ax-ie2 1508 ax-8 1518 ax-4 1524 ax-17 1540 ax-i9 1544 ax-ial 1548 ax-14 2170 ax-ext 2178 |
| This theorem depends on definitions: df-bi 117 df-nf 1475 |
| This theorem is referenced by: nninfinf 10535 |
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