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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 2019. (Contributed by NM, 14-Nov-2008.) |
Ref | Expression |
---|---|
axext4 | ⊢ (x = y ↔ ∀z(z ∈ x ↔ z ∈ y)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | elequ2 1598 | . . 3 ⊢ (x = y → (z ∈ x ↔ z ∈ y)) | |
2 | 1 | alrimiv 1751 | . 2 ⊢ (x = y → ∀z(z ∈ x ↔ z ∈ y)) |
3 | axext3 2020 | . 2 ⊢ (∀z(z ∈ x ↔ z ∈ y) → x = y) | |
4 | 2, 3 | impbii 117 | 1 ⊢ (x = y ↔ ∀z(z ∈ x ↔ z ∈ y)) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 98 ∀wal 1240 |
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-ie1 1379 ax-ie2 1380 ax-8 1392 ax-4 1397 ax-14 1402 ax-17 1416 ax-i9 1420 ax-ial 1424 ax-ext 2019 |
This theorem depends on definitions: df-bi 110 df-nf 1347 |
This theorem is referenced by: (None) |
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