Timeline for I am stuck on how to prove the contradiction of R(b,a) can anybody help me?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 26, 2018 at 13:30 | history | edited | Frank Hubeny | CC BY-SA 4.0 |
replace forall with universal
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| Jul 26, 2018 at 4:41 | comment | added | Frank Hubeny | @GrahamKemp I think the ~I would work. It would remove one more line. | |
| Jul 26, 2018 at 4:00 | comment | added | Graham Kemp | That checker explicitly lists ~E and ~I rules in the Basic Rules column. You can surely use them. | |
| Jul 26, 2018 at 3:36 | comment | added | Frank Hubeny | @GrahamKemp Since I am using a specific proof checker I have to follow those built in rules of that software. I can't start with Rba and deduce ~Rba. I have to start with the negation of what I want which would be ~~Rba. Regardless, it is still an indirect proof. | |
| Jul 26, 2018 at 3:03 | comment | added | Graham Kemp | A proof by negation is all you need. Assume Rba , derive the contradition, then use negation introduction to deduce ¬Rba . | |
| Jul 26, 2018 at 2:26 | comment | added | Frank Hubeny | @GrahamKemp Your suggestion worked. | |
| Jul 26, 2018 at 2:25 | history | edited | Frank Hubeny | CC BY-SA 4.0 |
added alternate proof
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| Jul 26, 2018 at 0:31 | comment | added | Frank Hubeny | @GrahamKemp I'll see if I can get a simpler proof with your suggestion. | |
| Jul 26, 2018 at 0:26 | comment | added | Graham Kemp | Eh. Don't mess with Modus Tollens and deMorgan's. Just use the core rules of inference: assume Rba, to derive Rab ˄ Rba and thus Raa, which contradicts ¬Raa so therefore deducing ¬Rba. | |
| Jul 25, 2018 at 20:29 | history | answered | Frank Hubeny | CC BY-SA 4.0 |