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Mirrors > Home > ILE Home > Th. List > nfcdeq | GIF version |
Description: If we have a conditional equality proof, where 𝜑 is 𝜑(𝑥) and 𝜓 is 𝜑(𝑦), and 𝜑(𝑥) in fact does not have 𝑥 free in it according to Ⅎ, then 𝜑(𝑥) ↔ 𝜑(𝑦) unconditionally. This proves that Ⅎ𝑥𝜑 is actually a not-free predicate. (Contributed by Mario Carneiro, 11-Aug-2016.) |
Ref | Expression |
---|---|
nfcdeq.1 | ⊢ Ⅎ𝑥𝜑 |
nfcdeq.2 | ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
Ref | Expression |
---|---|
nfcdeq | ⊢ (𝜑 ↔ 𝜓) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | nfcdeq.1 | . . 3 ⊢ Ⅎ𝑥𝜑 | |
2 | 1 | sbf 1764 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑) |
3 | nfv 1515 | . . 3 ⊢ Ⅎ𝑥𝜓 | |
4 | nfcdeq.2 | . . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) | |
5 | 4 | cdeqri 2932 | . . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
6 | 3, 5 | sbie 1778 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
7 | 2, 6 | bitr3i 185 | 1 ⊢ (𝜑 ↔ 𝜓) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 104 Ⅎwnf 1447 [wsb 1749 CondEqwcdeq 2929 |
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 1434 ax-gen 1436 ax-ie1 1480 ax-ie2 1481 ax-4 1497 ax-17 1513 ax-i9 1517 ax-ial 1521 |
This theorem depends on definitions: df-bi 116 df-nf 1448 df-sb 1750 df-cdeq 2930 |
This theorem is referenced by: nfccdeq 2944 |
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