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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 1788 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑) |
3 | nfv 1539 | . . 3 ⊢ Ⅎ𝑥𝜓 | |
4 | nfcdeq.2 | . . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) | |
5 | 4 | cdeqri 2963 | . . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
6 | 3, 5 | sbie 1802 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
7 | 2, 6 | bitr3i 186 | 1 ⊢ (𝜑 ↔ 𝜓) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 105 Ⅎwnf 1471 [wsb 1773 CondEqwcdeq 2960 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-5 1458 ax-gen 1460 ax-ie1 1504 ax-ie2 1505 ax-4 1521 ax-17 1537 ax-i9 1541 ax-ial 1545 |
This theorem depends on definitions: df-bi 117 df-nf 1472 df-sb 1774 df-cdeq 2961 |
This theorem is referenced by: nfccdeq 2975 |
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