Mixing
Discrete mathematics - logical equivalence?
$begingroup$
I'm asked to find a logical expression that is equivalent to the one listed in the question below, but I'm stumped as to what steps I would take next. If someone could show me step by step how to solve them and what rules would be used, I would really appreciate it.
- Using only the NOT and the AND operators, find an expression that is equivalent to ¬(a ∧ ¬b)↔𝑐
The furthest I got is (¬a∨b)↔c by using De Morgan's law, but I don't know how I can simplify it further than that. Someone told me the final answer is b ↔ c but I have no idea how they got there or if it's even correct?
discrete-mathematics logic propositional-calculus
edited Jan 27 at 6:11
Misha Lavrov
47.8k657107
asked Jan 26 at 20:33
BlackthornBlackthorn
316
$endgroup$
add a comment |
$begingroup$
I'm asked to find a logical expression that is equivalent to the one listed in the question below, but I'm stumped as to what steps I would take next. If someone could show me step by step how to solve them and what rules would be used, I would really appreciate it.
- Using only the NOT and the AND operators, find an expression that is equivalent to ¬(a ∧ ¬b)↔𝑐
The furthest I got is (¬a∨b)↔c by using De Morgan's law, but I don't know how I can simplify it further than that. Someone told me the final answer is b ↔ c but I have no idea how they got there or if it's even correct?
discrete-mathematics logic propositional-calculus
edited Jan 27 at 6:11
Misha Lavrov
47.8k657107
asked Jan 26 at 20:33
BlackthornBlackthorn
316
$endgroup$
$begingroup$
In part A) $bleftrightarrow c$ is incorrect. What if $a$ and $b$ are flase and $c$ is true?
$endgroup$
– saulspatz
Jan 26 at 21:30
$begingroup$
You want to remember that the biconditional can be broken up into two conditional statements with the AND: p $leftrightarrow$ q $equiv$ (p $rightarrow$q) $land$ (q $rightarrow$ p). Then you use the implication equivalence and keep going.
$endgroup$
– Joel Pereira
Jan 26 at 21:39
add a comment |
$begingroup$
I'm asked to find a logical expression that is equivalent to the one listed in the question below, but I'm stumped as to what steps I would take next. If someone could show me step by step how to solve them and what rules would be used, I would really appreciate it.
- Using only the NOT and the AND operators, find an expression that is equivalent to ¬(a ∧ ¬b)↔𝑐
The furthest I got is (¬a∨b)↔c by using De Morgan's law, but I don't know how I can simplify it further than that. Someone told me the final answer is b ↔ c but I have no idea how they got there or if it's even correct?
discrete-mathematics logic propositional-calculus
edited Jan 27 at 6:11
Misha Lavrov
47.8k657107
asked Jan 26 at 20:33
BlackthornBlackthorn
316
$endgroup$
I'm asked to find a logical expression that is equivalent to the one listed in the question below, but I'm stumped as to what steps I would take next. If someone could show me step by step how to solve them and what rules would be used, I would really appreciate it.
- Using only the NOT and the AND operators, find an expression that is equivalent to ¬(a ∧ ¬b)↔𝑐
The furthest I got is (¬a∨b)↔c by using De Morgan's law, but I don't know how I can simplify it further than that. Someone told me the final answer is b ↔ c but I have no idea how they got there or if it's even correct?
discrete-mathematics logic propositional-calculus
discrete-mathematics logic propositional-calculus
edited Jan 27 at 6:11
Misha Lavrov
47.8k657107
asked Jan 26 at 20:33
BlackthornBlackthorn
316
edited Jan 27 at 6:11
Misha Lavrov
47.8k657107
asked Jan 26 at 20:33
BlackthornBlackthorn
316
edited Jan 27 at 6:11
Misha Lavrov
47.8k657107
edited Jan 27 at 6:11
Misha Lavrov
47.8k657107
edited Jan 27 at 6:11
Misha Lavrov
47.8k657107
47.8k657107
asked Jan 26 at 20:33
BlackthornBlackthorn
316
asked Jan 26 at 20:33
BlackthornBlackthorn
316
asked Jan 26 at 20:33
BlackthornBlackthorn
316
316
$begingroup$
In part A) $bleftrightarrow c$ is incorrect. What if $a$ and $b$ are flase and $c$ is true?
$endgroup$
– saulspatz
Jan 26 at 21:30
$begingroup$
You want to remember that the biconditional can be broken up into two conditional statements with the AND: p $leftrightarrow$ q $equiv$ (p $rightarrow$q) $land$ (q $rightarrow$ p). Then you use the implication equivalence and keep going.
$endgroup$
– Joel Pereira
Jan 26 at 21:39
add a comment |
$begingroup$
In part A) $bleftrightarrow c$ is incorrect. What if $a$ and $b$ are flase and $c$ is true?
$endgroup$
– saulspatz
Jan 26 at 21:30
$begingroup$
You want to remember that the biconditional can be broken up into two conditional statements with the AND: p $leftrightarrow$ q $equiv$ (p $rightarrow$q) $land$ (q $rightarrow$ p). Then you use the implication equivalence and keep going.
$endgroup$
– Joel Pereira
Jan 26 at 21:39
$begingroup$
In part A) $bleftrightarrow c$ is incorrect. What if $a$ and $b$ are flase and $c$ is true?
$endgroup$
– saulspatz
Jan 26 at 21:30
$begingroup$
In part A) $bleftrightarrow c$ is incorrect. What if $a$ and $b$ are flase and $c$ is true?
$endgroup$
– saulspatz
Jan 26 at 21:30
$begingroup$
You want to remember that the biconditional can be broken up into two conditional statements with the AND: p $leftrightarrow$ q $equiv$ (p $rightarrow$q) $land$ (q $rightarrow$ p). Then you use the implication equivalence and keep going.
$endgroup$
– Joel Pereira
Jan 26 at 21:39
$begingroup$
You want to remember that the biconditional can be broken up into two conditional statements with the AND: p $leftrightarrow$ q $equiv$ (p $rightarrow$q) $land$ (q $rightarrow$ p). Then you use the implication equivalence and keep going.
$endgroup$
– Joel Pereira
Jan 26 at 21:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $A leftrightarrow B$ is logically equivalent to $(lnot A lor B) land (lnot B lor A)$, we have:
begin{align}
lnot (a land lnot b) leftrightarrow c &equiv ( lnot lnot (a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land lnot (c land a land lnot b) \
&equiv ((a lor c) land (lnot b lor c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (lnot lnot b land lnot c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (b land lnot c)) land lnot (c land a land lnot b) \
end{align}
answered Jan 26 at 23:12
Ruggiero RilieviRuggiero Rilievi
179112
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
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active
oldest
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active
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$begingroup$
Since $A leftrightarrow B$ is logically equivalent to $(lnot A lor B) land (lnot B lor A)$, we have:
begin{align}
lnot (a land lnot b) leftrightarrow c &equiv ( lnot lnot (a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land lnot (c land a land lnot b) \
&equiv ((a lor c) land (lnot b lor c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (lnot lnot b land lnot c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (b land lnot c)) land lnot (c land a land lnot b) \
end{align}
answered Jan 26 at 23:12
Ruggiero RilieviRuggiero Rilievi
179112
$endgroup$
add a comment |
$begingroup$
Since $A leftrightarrow B$ is logically equivalent to $(lnot A lor B) land (lnot B lor A)$, we have:
begin{align}
lnot (a land lnot b) leftrightarrow c &equiv ( lnot lnot (a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land lnot (c land a land lnot b) \
&equiv ((a lor c) land (lnot b lor c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (lnot lnot b land lnot c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (b land lnot c)) land lnot (c land a land lnot b) \
end{align}
answered Jan 26 at 23:12
Ruggiero RilieviRuggiero Rilievi
179112
$endgroup$
add a comment |
$begingroup$
Since $A leftrightarrow B$ is logically equivalent to $(lnot A lor B) land (lnot B lor A)$, we have:
begin{align}
lnot (a land lnot b) leftrightarrow c &equiv ( lnot lnot (a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land lnot (c land a land lnot b) \
&equiv ((a lor c) land (lnot b lor c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (lnot lnot b land lnot c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (b land lnot c)) land lnot (c land a land lnot b) \
end{align}
answered Jan 26 at 23:12
Ruggiero RilieviRuggiero Rilievi
179112
$endgroup$
Since $A leftrightarrow B$ is logically equivalent to $(lnot A lor B) land (lnot B lor A)$, we have:
begin{align}
lnot (a land lnot b) leftrightarrow c &equiv ( lnot lnot (a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land (lnot c lor lnot (a land lnot b)) \
&equiv ((a land lnot b) lor c) land lnot (c land a land lnot b) \
&equiv ((a lor c) land (lnot b lor c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (lnot lnot b land lnot c)) land lnot (c land a land lnot b) \
&equiv (lnot (lnot a land lnot c) land lnot (b land lnot c)) land lnot (c land a land lnot b) \
end{align}
answered Jan 26 at 23:12
Ruggiero RilieviRuggiero Rilievi
179112
answered Jan 26 at 23:12
Ruggiero RilieviRuggiero Rilievi
179112
answered Jan 26 at 23:12
Ruggiero RilieviRuggiero Rilievi
179112
answered Jan 26 at 23:12
Ruggiero RilieviRuggiero Rilievi
179112
179112
add a comment |
add a comment |
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$begingroup$
In part A) $bleftrightarrow c$ is incorrect. What if $a$ and $b$ are flase and $c$ is true?
$endgroup$
– saulspatz
Jan 26 at 21:30
$begingroup$
You want to remember that the biconditional can be broken up into two conditional statements with the AND: p $leftrightarrow$ q $equiv$ (p $rightarrow$q) $land$ (q $rightarrow$ p). Then you use the implication equivalence and keep going.
$endgroup$
– Joel Pereira
Jan 26 at 21:39