// Numbas version: finer_feedback_settings
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Use the truth tables for the \"and\" operator $\\wedge$ and the implication $\\Rightarrow$. Remember that $p\\Rightarrow q$ is true whenever $p$ is false or $q$ is true.
\n", "rulesets": {}, "parts": [{"prompt": "Complete the following truth table; enter $0$ for false and $1$ for true.
\n\n\n$p$ | $q$ | $r$ | $q\\land r$ | $p\\Rightarrow(q\\land r)$ | $p\\Rightarrow q$ | $p\\Rightarrow r$ | $(p\\Rightarrow q)\\land(p\\Rightarrow r)$ |
\n\n0 | \n0 | \n0 | \n[[0]] | \n[[8]] | \n[[16]] | \n[[24]] | \n[[32]] | \n
\n\n0 | \n0 | \n1 | \n[[1]] | \n[[9]] | \n[[17]] | \n[[25]] | \n[[33]] | \n
\n\n0 | \n1 | \n0 | \n[[2]] | \n[[10]] | \n[[18]] | \n[[26]] | \n[[34]] | \n
\n\n0 | \n1 | \n1 | \n[[3]] | \n[[11]] | \n[[19]] | \n[[27]] | \n[[35]] | \n
\n\n1 | \n0 | \n0 | \n[[4]] | \n[[12]] | \n[[20]] | \n[[28]] | \n[[36]] | \n
\n\n1 | \n0 | \n1 | \n[[5]] | \n[[13]] | \n[[21]] | \n[[29]] | \n[[37]] | \n
\n\n1 | \n1 | \n0 | \n[[6]] | \n[[14]] | \n[[22]] | \n[[30]] | \n[[38]] | \n
\n\n1 | \n1 | \n1 | \n[[7]] | \n[[15]] | \n[[23]] | \n[[31]] | \n[[39]] | \n
\n\n
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expressions $p\\Rightarrow(q\\land r)$ and $(p\\Rightarrow q)\\land (p\\Rightarrow r)$ logically equivalent?
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\n$p\\Rightarrow(q\\wedge r)$ and $(p\\Rightarrow q)\\wedge (p\\Rightarrow r)$, using all possible combinations of truth values for $p,q,r$.
\nRecall that $0$ means false and $1$ means true.
\n\n", "variable_groups": [], "variablesTest": {"maxRuns": "150", "condition": ""}, "variables": {}, "metadata": {"description": "Create a truth table for a logical expression of the form $(a \\operatorname{op1} b) \\operatorname{op2}(c \\operatorname{op3} d)$ where $a, \\;b,\\;c,\\;d$ can be the Boolean variables $p,\\;q,\\;\\neg p,\\;\\neg q$ and each of $\\operatorname{op1},\\;\\operatorname{op2},\\;\\operatorname{op3}$ one of $\\lor,\\;\\land,\\;\\to$.
\nFor example: $(p \\lor \\neg q) \\land(q \\to \\neg p)$.
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