Timeline for How to prove that "n number of proposition(s) give rise to 2^(n) number of truth-value combinations" using mathematical induction?
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7 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2018 at 11:59 | comment | added | ancient mathematician | I am sorry, I thought you wanted the number of what you were calling "complex propositions". | |
| Dec 20, 2018 at 11:41 | vote | accept | user628101 | ||
| Dec 20, 2018 at 11:18 | answer | added | Maged Saeed | timeline score: 0 | |
| Dec 20, 2018 at 11:18 | comment | added | user628101 | Certainly not! When there are 3 atomic propositions, the combinations are: TTT, FFF, TTF, TFT, FTT, FFT, TFF, FTF, which is 2^3 = 8 propositions. | |
| Dec 20, 2018 at 11:12 | comment | added | ancient mathematician | I think you mean $2^{2^n}$. When $n=1$ there are four functions $f(x)$: (i) $f(x)=T$ for all $x$; (ii) $f(x)=F$ for all $x$; (iii) $f(x)=x$ for all $x$; (iv) $f(x)=not-x$ for all $x$. | |
| Dec 20, 2018 at 11:05 | review | First posts | |||
| Dec 20, 2018 at 11:06 | |||||
| Dec 20, 2018 at 11:02 | history | asked | user628101 | CC BY-SA 4.0 |