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Mirrors > Home > MPE Home > Th. List > nfcdeq | Structured version Visualization version 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 2368 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑) |
3 | nfv 1830 | . . 3 ⊢ Ⅎ𝑥𝜓 | |
4 | nfcdeq.2 | . . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) | |
5 | 4 | cdeqri 3388 | . . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
6 | 3, 5 | sbie 2396 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
7 | 2, 6 | bitr3i 265 | 1 ⊢ (𝜑 ↔ 𝜓) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 195 Ⅎwnf 1699 [wsb 1867 CondEqwcdeq 3385 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-12 2034 ax-13 2234 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-cdeq 3386 |
This theorem is referenced by: nfccdeq 3400 |
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