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Mirrors > Home > ILE Home > Th. List > cdeqab1 | GIF version |
Description: Distribute conditional equality over abstraction. (Contributed by Mario Carneiro, 11-Aug-2016.) |
Ref | Expression |
---|---|
cdeqnot.1 | ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
Ref | Expression |
---|---|
cdeqab1 | ⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓}) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | cdeqnot.1 | . . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) | |
2 | 1 | cdeqri 2812 | . . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
3 | 2 | cbvabv 2206 | . 2 ⊢ {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓} |
4 | 3 | cdeqth 2813 | 1 ⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓}) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 103 = wceq 1285 {cab 2069 CondEqwcdeq 2809 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-io 663 ax-5 1377 ax-7 1378 ax-gen 1379 ax-ie1 1423 ax-ie2 1424 ax-8 1436 ax-10 1437 ax-11 1438 ax-i12 1439 ax-bndl 1440 ax-4 1441 ax-17 1460 ax-i9 1464 ax-ial 1468 ax-i5r 1469 ax-ext 2065 |
This theorem depends on definitions: df-bi 115 df-nf 1391 df-sb 1688 df-clab 2070 df-cleq 2076 df-cdeq 2810 |
This theorem is referenced by: (None) |
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