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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 2119. (Contributed by NM, 14-Nov-2008.) |
Ref | Expression |
---|---|
axext4 | ⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | elequ2 1691 | . . 3 ⊢ (𝑥 = 𝑦 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) | |
2 | 1 | alrimiv 1846 | . 2 ⊢ (𝑥 = 𝑦 → ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) |
3 | axext3 2120 | . 2 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦) | |
4 | 2, 3 | impbii 125 | 1 ⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦)) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 104 ∀wal 1329 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-5 1423 ax-gen 1425 ax-ie1 1469 ax-ie2 1470 ax-8 1482 ax-4 1487 ax-14 1492 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-ext 2119 |
This theorem depends on definitions: df-bi 116 df-nf 1437 |
This theorem is referenced by: (None) |
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