Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
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 2390. (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 2271 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑) |
3 | nfv 1915 | . . 3 ⊢ Ⅎ𝑥𝜓 | |
4 | nfcdeq.2 | . . . 4 ⊢ CondEq(𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) | |
5 | 4 | cdeqri 3759 | . . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) |
6 | 3, 5 | sbie 2544 | . 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓) |
7 | 2, 6 | bitr3i 279 | 1 ⊢ (𝜑 ↔ 𝜓) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 208 Ⅎwnf 1784 [wsb 2069 CondEqwcdeq 3756 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-10 2145 ax-12 2177 ax-13 2390 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-ex 1781 df-nf 1785 df-sb 2070 df-cdeq 3757 |
This theorem is referenced by: nfccdeq 3771 |
Copyright terms: Public domain | W3C validator |