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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. Usage of this theorem is discouraged because it depends on ax-13 2374. (Contributed by Mario Carneiro, 11-Aug-2016.) (New usage is discouraged.) |
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 2267 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑) |
3 | nfv 1921 | . . 3 ⊢ Ⅎ𝑥𝜓 | |
4 | nfcdeq.2 | . . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) | |
5 | 4 | cdeqri 3705 | . . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
6 | 3, 5 | sbie 2508 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
7 | 2, 6 | bitr3i 276 | 1 ⊢ (𝜑 ↔ 𝜓) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 205 Ⅎwnf 1790 [wsb 2071 CondEqwcdeq 3702 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1802 ax-4 1816 ax-5 1917 ax-6 1975 ax-7 2015 ax-10 2141 ax-12 2175 ax-13 2374 |
This theorem depends on definitions: df-bi 206 df-an 397 df-or 845 df-ex 1787 df-nf 1791 df-sb 2072 df-cdeq 3703 |
This theorem is referenced by: nfccdeq 3717 |
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