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#1
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how to build a "theorem prover" in Excel
My objective is to teach formal logic (some theorems of propositional logic)
to people who may know the Excel basic functions and have some knowledge of logic. My idea is to use a very simple true table, so I can give it in a beginners class. Example : I want to prove the formula : (A^ B) - B 1st STEP (done!) column A is associated to the function "A", in Excel column B is associated to the function "B", in Excel columns A, B., have the logic values of A and B column C has the formula to prove TABLE A B (A^ B ) - B 1 0 0 = OR ((A*B) <= B) +0 2 1 0 = OR ((A*B) <= B) +0 3 0 1 = OR ((A*B) <= B) +0 4 1 1 = OR ((A*B) <= B) +0 result A B (A^ B ) - B 1 0 0 1 2 1 0 1 3 0 1 1 4 1 1 1 So the given formula is true for all the values of A and B. 2nd STEP It is given the formula (A^ B) - B written in cell C1, and the true table must verify it automatically. I tried the text functions, but it does not work A B C 1 (A*B ) =< B 2 0 0 = ? 3 1 0 = ? 4 0 1 = ? 5 1 1 = ? expected result A B C 1 (A*B ) =< B 2 0 0 1 3 1 0 1 4 0 1 1 5 1 1 1 Any ideas ? sorry for my English Regards Alvaro |
#2
Posted to microsoft.public.excel.worksheet.functions
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how to build a "theorem prover" in Excel
I cannot recall how to interpret (A^B)-B
Is it A implies B, therefore if A is true B is true? On to Excel. Why not use TRUE and FALSE in place of 1/0 But if you do want 1/10 then use =-- OR ((A*B) <= B) This gets rid of the confusing + 0. Right after the = we have two negations to coerce Boolean to numeric. best wishes from Archimedes -- Bernard Liengme Microsoft Excel MVP http://people.stfx.ca/bliengme "socrates" wrote in message ... My objective is to teach formal logic (some theorems of propositional logic) to people who may know the Excel basic functions and have some knowledge of logic. My idea is to use a very simple true table, so I can give it in a beginners class. Example : I want to prove the formula : (A^ B) - B 1st STEP (done!) column A is associated to the function "A", in Excel column B is associated to the function "B", in Excel columns A, B., have the logic values of A and B column C has the formula to prove TABLE A B (A^ B ) - B 1 0 0 = OR ((A*B) <= B) +0 2 1 0 = OR ((A*B) <= B) +0 3 0 1 = OR ((A*B) <= B) +0 4 1 1 = OR ((A*B) <= B) +0 result A B (A^ B ) - B 1 0 0 1 2 1 0 1 3 0 1 1 4 1 1 1 So the given formula is true for all the values of A and B. 2nd STEP It is given the formula (A^ B) - B written in cell C1, and the true table must verify it automatically. I tried the text functions, but it does not work A B C 1 (A*B ) =< B 2 0 0 = ? 3 1 0 = ? 4 0 1 = ? 5 1 1 = ? expected result A B C 1 (A*B ) =< B 2 0 0 1 3 1 0 1 4 0 1 1 5 1 1 1 Any ideas ? sorry for my English Regards Alvaro |
#3
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how to build a "theorem prover" in Excel
"I cannot recall how to interpret (A^B)-B
(A^B)-B means: A and B, implies B The proof of this formula in proposition logic is (A^B) - B <= ~(A^B) v B <= ~A v ~B v B <= T The above formula is true whatever the logical values of A and B. Since ~B v B is true for whatever be the value of B "Is it A implies B, therefore if A is true B is true?" Let us analyse your logical expression : (A - B) - (A-B) Let us do C = (A -B) C represents the implication Then <= C - C <= ~C v C <= T your formula is true whatever the logical values of A and B What I want is to avoid to write the logical expression (in excel format) in all the lines of the true table For 2 variables is 4 lines For 3 variables is 8 lines ..... For 6 variables is 64 lines...! What I want is to write the formula in normal logic frame only once. Then the true table using any excel functon transforms the formula, and computes it in all the lines. If all the lines are "true" or 1, the formula is a tautology Regards Alvaro ************************************************** ********* I cannot recall how to interpret (A^B)-B Is it A implies B, therefore if A is true B is true? On to Excel. Why not use TRUE and FALSE in place of 1/0 But if you do want 1/10 then use =-- OR ((A*B) <= B) This gets rid of the confusing + 0. Right after the = we have two negations to coerce Boolean to numeric. best wishes from Archimedes -- Bernard Liengme Microsoft Excel MVP http://people.stfx.ca/bliengme "socrates" wrote in message ... My objective is to teach formal logic (some theorems of propositional logic) to people who may know the Excel basic functions and have some knowledge of logic. My idea is to use a very simple true table, so I can give it in a beginners class. Example : I want to prove the formula : (A^ B) - B 1st STEP (done!) column A is associated to the function "A", in Excel column B is associated to the function "B", in Excel columns A, B., have the logic values of A and B column C has the formula to prove TABLE A B (A^ B ) - B 1 0 0 = OR ((A*B) <= B) +0 2 1 0 = OR ((A*B) <= B) +0 3 0 1 = OR ((A*B) <= B) +0 4 1 1 = OR ((A*B) <= B) +0 result A B (A^ B ) - B 1 0 0 1 2 1 0 1 3 0 1 1 4 1 1 1 So the given formula is true for all the values of A and B. 2nd STEP It is given the formula (A^ B) - B written in cell C1, and the true table must verify it automatically. I tried the text functions, but it does not work A B C 1 (A*B ) =< B 2 0 0 = ? 3 1 0 = ? 4 0 1 = ? 5 1 1 = ? expected result A B C 1 (A*B ) =< B 2 0 0 1 3 1 0 1 4 0 1 1 5 1 1 1 Any ideas ? sorry for my English Regards Alvaro |
#4
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how to build a "theorem prover" in Excel
A B (A^ B ) - B
As a side note, it's a little easier in vba. Fx = (a And b) Imp b -- Dana DeLouis "alvaro" wrote in message ... "I cannot recall how to interpret (A^B)-B (A^B)-B means: A and B, implies B The proof of this formula in proposition logic is (A^B) - B <= ~(A^B) v B <= ~A v ~B v B <= T The above formula is true whatever the logical values of A and B. Since ~B v B is true for whatever be the value of B "Is it A implies B, therefore if A is true B is true?" Let us analyse your logical expression : (A - B) - (A-B) Let us do C = (A -B) C represents the implication Then <= C - C <= ~C v C <= T your formula is true whatever the logical values of A and B What I want is to avoid to write the logical expression (in excel format) in all the lines of the true table For 2 variables is 4 lines For 3 variables is 8 lines .... For 6 variables is 64 lines...! What I want is to write the formula in normal logic frame only once. Then the true table using any excel functon transforms the formula, and computes it in all the lines. If all the lines are "true" or 1, the formula is a tautology Regards Alvaro ************************************************** ********* I cannot recall how to interpret (A^B)-B Is it A implies B, therefore if A is true B is true? On to Excel. Why not use TRUE and FALSE in place of 1/0 But if you do want 1/10 then use =-- OR ((A*B) <= B) This gets rid of the confusing + 0. Right after the = we have two negations to coerce Boolean to numeric. best wishes from Archimedes -- Bernard Liengme Microsoft Excel MVP http://people.stfx.ca/bliengme "socrates" wrote in message ... My objective is to teach formal logic (some theorems of propositional logic) to people who may know the Excel basic functions and have some knowledge of logic. My idea is to use a very simple true table, so I can give it in a beginners class. Example : I want to prove the formula : (A^ B) - B 1st STEP (done!) column A is associated to the function "A", in Excel column B is associated to the function "B", in Excel columns A, B., have the logic values of A and B column C has the formula to prove TABLE A B (A^ B ) - B 1 0 0 = OR ((A*B) <= B) +0 2 1 0 = OR ((A*B) <= B) +0 3 0 1 = OR ((A*B) <= B) +0 4 1 1 = OR ((A*B) <= B) +0 result A B (A^ B ) - B 1 0 0 1 2 1 0 1 3 0 1 1 4 1 1 1 So the given formula is true for all the values of A and B. 2nd STEP It is given the formula (A^ B) - B written in cell C1, and the true table must verify it automatically. I tried the text functions, but it does not work A B C 1 (A*B ) =< B 2 0 0 = ? 3 1 0 = ? 4 0 1 = ? 5 1 1 = ? expected result A B C 1 (A*B ) =< B 2 0 0 1 3 1 0 1 4 0 1 1 5 1 1 1 Any ideas ? sorry for my English Regards Alvaro |
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