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Mirrors > Home > ILE Home > Th. List > nfcdeq | Unicode version |
Description: If we have a conditional equality proof, where is and is , and in fact does not have free in it according to , then unconditionally. This proves that is actually a not-free predicate. (Contributed by Mario Carneiro, 11-Aug-2016.) |
Ref | Expression |
---|---|
nfcdeq.1 | |
nfcdeq.2 | CondEq |
Ref | Expression |
---|---|
nfcdeq |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | nfcdeq.1 | . . 3 | |
2 | 1 | sbf 1660 | . 2 |
3 | nfv 1421 | . . 3 | |
4 | nfcdeq.2 | . . . 4 CondEq | |
5 | 4 | cdeqri 2750 | . . 3 |
6 | 3, 5 | sbie 1674 | . 2 |
7 | 2, 6 | bitr3i 175 | 1 |
Colors of variables: wff set class |
Syntax hints: wb 98 wnf 1349 wsb 1645 CondEqwcdeq 2747 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1336 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-4 1400 ax-17 1419 ax-i9 1423 ax-ial 1427 |
This theorem depends on definitions: df-bi 110 df-nf 1350 df-sb 1646 df-cdeq 2748 |
This theorem is referenced by: nfccdeq 2762 |
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