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Create a truth table for a logical expression of the form (aop1b)op2(cop3d) where a,b,c,d can be the Boolean variables p,q,¬p,¬q and each of op1,op2,op3 one of ∨,∧,→.
For example: (p∨¬q)∧(q→¬p).
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England schools
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England university
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Scotland schools
Taxonomy: mathcentre
Taxonomy: Kind of activity
Taxonomy: Context
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Bernhard von Stengel 8 years, 2 months ago
Created this as a copy of Truth tables 1(v2).There are 18 other versions that do you not have access to.
Name | Type | Generated Value |
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logic_symbol_list | list |
[ "p", "q", "not p", "not q" ]
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latex_symbol_list | list |
List of 4 items
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s | list |
[ 1, 2, 1, 0 ]
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Name | Type | Generated Value |
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a | string |
q
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b | string |
\neg p
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op | string |
\lor
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pre_ev1 | list |
[ true, false, true, true ]
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ev1 | list |
[ "T", "F", "T", "T" ]
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Name | Type | Generated Value |
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a1 | string |
q
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b1 | string |
p
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op2 | string |
\lor
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pre_ev2 | list |
[ true, true, true, false ]
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ev2 | list |
[ "T", "T", "T", "F" ]
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Name | Type | Generated Value |
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p | list |
[ true, true, false, false ]
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q | list |
[ true, false, true, false ]
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disp | list |
[ "T", "T", "F", "F" ]
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disq | list |
[ "T", "F", "T", "F" ]
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Name | Type | Generated Value |
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op1 | string |
\land
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t_value | list |
[ "T", "F", "T", "F" ]
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Generated value: list
[ "p", "q", "not p", "not q" ]
This variable doesn't seem to be used anywhere.
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Complete the following truth table:
p | q | {a}{op}{b} | {a1}{op2}{b1} | ({a}{op}{b}){op1}({a1}{op2}{b1}) |
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{disp[0]} | {disq[0]} | |||
{disp[1]} | {disq[1]} | |||
{disp[2]} | {disq[2]} | |||
{disp[3]} | {disq[3]} |
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