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Author Question: By applying the rule Bicond to (M F) (E L) we can deduce (Read 31 times)

effaey

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on: May 12, 2022

Question 1

By applying the rule Bicond to “(E ∙ L) v (~ E ∙ ~ L)” we can deduce
◦ ~ E ∙ ~ L
◦ E ∙ L
◦ E ≡ L

Question 2

By applying the rule Bicond to “(M ∙ F) ≡ (E L)” we can deduce
◦ [(M ∙ F) ⊃ (E ∙ L)] v [(E ∙ L) ⊃ (M ∙ F)]
◦ [(M ∙ F) ⊃ (E ∙ L)] ∙ [(E ∙ L) ⊃ (M ∙ F)]
◦ [(M ∙ F) ⊃ (E ∙ L)]
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Marked as best answer by effaey on May 12, 2022

nikspatel037

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Reply #1 on: May 12, 2022
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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effaey

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Reply 2 on: May 12, 2022
Excellent

EAN94

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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