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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 2899 | . . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
3 | 2 | cbvabv 2265 | . 2 ⊢ {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓} |
4 | 3 | cdeqth 2900 | 1 ⊢ CondEq(𝑥 = 𝑦 → {𝑥 ∣ 𝜑} = {𝑦 ∣ 𝜓}) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 104 = wceq 1332 {cab 2126 CondEqwcdeq 2896 |
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-io 699 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1483 ax-10 1484 ax-11 1485 ax-i12 1486 ax-bndl 1487 ax-4 1488 ax-17 1507 ax-i9 1511 ax-ial 1515 ax-i5r 1516 ax-ext 2122 |
This theorem depends on definitions: df-bi 116 df-nf 1438 df-sb 1737 df-clab 2127 df-cleq 2133 df-cdeq 2897 |
This theorem is referenced by: (None) |
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