document.write( "Question 664426: Given Ź(p∧q), prove (Źp∨Źq). \n" ); document.write( "
Algebra.Com's Answer #413333 by Edwin McCravy(20055) $\"\"$ $\"About$
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document.write( "Ź(p∧q), prove (Źp∨Źq).\r\n" );
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document.write( "This is done by a truth table.\r\n" );
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document.write( "1. Negation:  Rule for Źx is: If x has a T, then Źx has\r\n" );
document.write( "   an F and vice-versa.\r\n" );
document.write( "2. Conjunction: Rule for x∧y is: If x and y both have\r\n" );
document.write( "   T's, x∧y has T, otherwise x∧y has F.  \r\n" );
document.write( "3. Disjunction: Rule for x∨y is: If x and y both have \r\n" );
document.write( "   F's, x∨y has F, otherwise x∨y has T.\r\n" );
document.write( "4. Biconditional:  Rule for x<->y is: If both x and y \r\n" );
document.write( "   have the same truth value, x<->y has T, otherwise F. \r\n" );
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document.write( "You need headings as listed below:   \r\n" );
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document.write( "p | q | Źp | Źq | p∧q | Ź(p∧q) | Źp∨Źq | Ź(p∧q) <-> (Źp∨Źq) |\r\n" );
document.write( "---------------------------------------------------------------\r\n" );
document.write( "T | T |  F |  F |  T  |    F    |   F   |         T           |\r\n" );
document.write( "T | F |    |    |     |         |       |                     |\r\n" );
document.write( "F | T |    |    |     |         |       |                     |\r\n" );
document.write( "F | F |    |    |     |         |       |                     |\r\n" );
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document.write( "The F under Źp is because there is a T under p, rule 1 above.\r\n" );
document.write( "The F under Źq is because there is a T under q, rule 1 above.\r\n" );
document.write( "The T under p∧q is because there is a T under both p and q,\r\n" );
document.write( "    rule 2 above.\r\n" );
document.write( "The F under Ź(p∧q) is because there is a T under p∧q,\r\n" );
document.write( "    rule 1 above.\r\n" );
document.write( "The F under Źp∨Źq is because Źp and Źq both have F's\r\n" );
document.write( "    rule 3 above.\r\n" );
document.write( "The T under Ź(p∧q) <-> (Źp∨Źq) is because both Ź(p∧q) and (Źp∨Źq)\r\n" );
document.write( "    have the same truth value F, rule 4\r\n" );
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document.write( "Now we fil in the next row:\r\n" );
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document.write( "p | q | Źp | Źq | p∧q | Ź(p∧q) | Źp∨Źq | Ź(p∧q) <-> (Źp∨Źq) |\r\n" );
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document.write( "T | T |  F |  F |  T  |    F    |   F   |         T           |\r\n" );
document.write( "T | F |  F |  T |  F  |    T    |   T   |         T           |\r\n" );
document.write( "F | T |    |    |     |         |       |                     |\r\n" );
document.write( "F | F |    |    |     |         |       |                     |\r\n" );
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document.write( "The F under Źp is because there is a T under p, rule 1 above.\r\n" );
document.write( "The T under Źq is because there is a F under q, rule 1 above.\r\n" );
document.write( "The F under p∧q is because p and q don't both have T's, rule 2 \r\n" );
document.write( "above.\r\n" );
document.write( "The T under Ź(p∧q) is because there is a F under p∧q,\r\n" );
document.write( "    rule 1 above.\r\n" );
document.write( "The T under Źp∨Źq is because Źp and Źq don't both have F's\r\n" );
document.write( "    rule 3 above.\r\n" );
document.write( "The T under Ź(p∧q) <-> (Źp∨Źq) is because both Ź(p∧q) and (Źp∨Źq)\r\n" );
document.write( "    have the same truth value T, rule 4\r\n" );
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document.write( "Now we fil in the third row:\r\n" );
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document.write( "p | q | Źp | Źq | p∧q | Ź(p∧q) | Źp∨Źq | Ź(p∧q) <-> (Źp∨Źq) |\r\n" );
document.write( "---------------------------------------------------------------\r\n" );
document.write( "T | T |  F |  F |  T  |    F    |   F   |         T           |\r\n" );
document.write( "T | F |  F |  T |  F  |    T    |   T   |         T           |\r\n" );
document.write( "F | T |  T |  F |  F  |    T    |   T   |         T           |\r\n" );
document.write( "F | F |    |    |     |         |       |                     |\r\n" );
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document.write( "The T under Źp is because there is a F under p, rule 1 above.\r\n" );
document.write( "The F under Źq is because there is a T under q, rule 1 above.\r\n" );
document.write( "The F under p∧q is because p and q don't both have T's, rule 2 \r\n" );
document.write( "above.\r\n" );
document.write( "The T under Ź(p∧q) is because there is a F under p∧q,\r\n" );
document.write( "    rule 1 above.\r\n" );
document.write( "The T under Źp∨Źq is because Źp and Źq don't both have F's\r\n" );
document.write( "    rule 3 above.\r\n" );
document.write( "The T under Ź(p∧q) <-> (Źp∨Źq) is because both Ź(p∧q) and (Źp∨Źq)\r\n" );
document.write( "    have the same truth value T, rule 4\r\n" );
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document.write( "Now we fill in the fourth row:\r\n" );
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document.write( "p | q | Źp | Źq | p∧q | Ź(p∧q) | Źp∨Źq | Ź(p∧q) <-> (Źp∨Źq) |\r\n" );
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document.write( "T | T |  F |  F |  T  |    F    |   F   |         T           |\r\n" );
document.write( "T | F |  F |  T |  F  |    T    |   T   |         T           |\r\n" );
document.write( "F | T |  T |  F |  F  |    T    |   T   |         T           |\r\n" );
document.write( "F | F |  T |  T |  F  |    T    |   T   |         T           |\r\n" );
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document.write( "The T under Źp is because there is a F under p, rule 1 above.\r\n" );
document.write( "The T under Źq is because there is a F under q, rule 1 above.\r\n" );
document.write( "The F under p∧q is because p and q don't both have T's, rule 2 \r\n" );
document.write( "above.\r\n" );
document.write( "The T under Ź(p∧q) is because there is a F under p∧q,\r\n" );
document.write( "    rule 1 above.\r\n" );
document.write( "The T under Źp∨Źq is because Źp and Źq don't both have F's\r\n" );
document.write( "    rule 3 above.\r\n" );
document.write( "The T under Ź(p∧q) <-> (Źp∨Źq) is because both Ź(p∧q) and (Źp∨Źq)\r\n" );
document.write( "    have the same truth value T, rule 4.\r\n" );
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document.write( "Now since there are only T's in the last column, that proves that\r\n" );
document.write( "in every case, the biconditional Ź(p∧q) <-> (Źp∨Źq) is\r\n" );
document.write( "true and therefore Ź(p∧q) and (Źp∨Źq) are equivalent\r\n" );
document.write( "because the biconditional is a tautology.\r\n" );
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document.write( "Edwin
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