Mixing
How to distribute $(neg 𝑃 lor 𝑅) land [((neg 𝑃 lor 𝑄) land (𝑃 lor neg 𝑄))∨((neg 𝑅 lor...
$begingroup$
I see two possible ways to distribute this.
I believe it is case 1.
For brevity,
Let
begin{align}A &= (neg P lor Q) &
B &= (P lor neg Q) & C &=(neg R lor Q) & D &=(R lor neg Q).
end{align}
Case 1:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg Pland A land B)lor(neg P land C land D)]lor [(R land A land B)lor (R land C land D)]$$
Case 2:
$$(neg P lor R)land [(A land B)lor (Cland D)]=$$
$$[(neg P lor A) land (neg P land B)lor(neg 𝑃 land C)∧(neg 𝑃 land D]lor [(R land A)land (R land B)) lor (R land C) land (R land D)]$$
Question: Which case is correct (if either)? Why is this case correct/other wrong?
logic propositional-calculus boolean-algebra
edited Feb 2 at 21:15
Taroccoesbrocco
5,87471840
asked Feb 2 at 20:26
tree_traversaltree_traversal
386
$endgroup$
add a comment |
$begingroup$
I see two possible ways to distribute this.
I believe it is case 1.
For brevity,
Let
begin{align}A &= (neg P lor Q) &
B &= (P lor neg Q) & C &=(neg R lor Q) & D &=(R lor neg Q).
end{align}
Case 1:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg Pland A land B)lor(neg P land C land D)]lor [(R land A land B)lor (R land C land D)]$$
Case 2:
$$(neg P lor R)land [(A land B)lor (Cland D)]=$$
$$[(neg P lor A) land (neg P land B)lor(neg 𝑃 land C)∧(neg 𝑃 land D]lor [(R land A)land (R land B)) lor (R land C) land (R land D)]$$
Question: Which case is correct (if either)? Why is this case correct/other wrong?
logic propositional-calculus boolean-algebra
edited Feb 2 at 21:15
Taroccoesbrocco
5,87471840
asked Feb 2 at 20:26
tree_traversaltree_traversal
386
$endgroup$
add a comment |
$begingroup$
I see two possible ways to distribute this.
I believe it is case 1.
For brevity,
Let
begin{align}A &= (neg P lor Q) &
B &= (P lor neg Q) & C &=(neg R lor Q) & D &=(R lor neg Q).
end{align}
Case 1:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg Pland A land B)lor(neg P land C land D)]lor [(R land A land B)lor (R land C land D)]$$
Case 2:
$$(neg P lor R)land [(A land B)lor (Cland D)]=$$
$$[(neg P lor A) land (neg P land B)lor(neg 𝑃 land C)∧(neg 𝑃 land D]lor [(R land A)land (R land B)) lor (R land C) land (R land D)]$$
Question: Which case is correct (if either)? Why is this case correct/other wrong?
logic propositional-calculus boolean-algebra
edited Feb 2 at 21:15
Taroccoesbrocco
5,87471840
asked Feb 2 at 20:26
tree_traversaltree_traversal
386
$endgroup$
I see two possible ways to distribute this.
I believe it is case 1.
For brevity,
Let
begin{align}A &= (neg P lor Q) &
B &= (P lor neg Q) & C &=(neg R lor Q) & D &=(R lor neg Q).
end{align}
Case 1:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg Pland A land B)lor(neg P land C land D)]lor [(R land A land B)lor (R land C land D)]$$
Case 2:
$$(neg P lor R)land [(A land B)lor (Cland D)]=$$
$$[(neg P lor A) land (neg P land B)lor(neg 𝑃 land C)∧(neg 𝑃 land D]lor [(R land A)land (R land B)) lor (R land C) land (R land D)]$$
Question: Which case is correct (if either)? Why is this case correct/other wrong?
logic propositional-calculus boolean-algebra
logic propositional-calculus boolean-algebra
edited Feb 2 at 21:15
Taroccoesbrocco
5,87471840
asked Feb 2 at 20:26
tree_traversaltree_traversal
386
edited Feb 2 at 21:15
Taroccoesbrocco
5,87471840
asked Feb 2 at 20:26
tree_traversaltree_traversal
386
edited Feb 2 at 21:15
Taroccoesbrocco
5,87471840
edited Feb 2 at 21:15
Taroccoesbrocco
5,87471840
edited Feb 2 at 21:15
Taroccoesbrocco
5,87471840
5,87471840
asked Feb 2 at 20:26
tree_traversaltree_traversal
386
asked Feb 2 at 20:26
tree_traversaltree_traversal
386
asked Feb 2 at 20:26
tree_traversaltree_traversal
386
386
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Case 1 is correct.
What you end up with in case 2 is ungrammatical ... you'll need to add some parentheses. I assume you meant:
$$[color{red}((neg P color{red}land A) land (neg P land B)color{red})lorcolor{red}((neg 𝑃 land C)∧(neg 𝑃 land Dcolor{red})color{red})]lor [color{red}((R land A)land (R land B)) lor color{red}((R land C) land (R land D)color{red})]$$
But note that that is equivalent to:
$$[((neg P land A) land (neg P land B))lor((neg 𝑃 land C)land(neg 𝑃 land D))]lor [((R land A)land (R land B)) lor ((R land C) land (R land D))] overset{Association (drop parentheses in generalized conjunctions)}Leftrightarrow$$
$$[(neg P land A land neg P land B)lor(neg 𝑃 land Cland neg 𝑃 land D)]lor [(R land Aland R land B) lor (R land C land R land D)] overset{Commutation}Leftrightarrow$$
$$[(neg P land neg P land A land B)lor(neg 𝑃 land neg 𝑃 land C land D)]lor [(R land R land A land B) lor (R land R land C land D)] overset{Idempotence}Leftrightarrow$$
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
... which is exactly what you got from case 1
In other words, had you used proper parentheses, both results would have been correct!
That said, I have a feeling you did the following to get to case 2:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg P land ((Aland B)lor (Cland D))] lor [(R land ((Aland B)lor (Cland D))]=$$
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]=$$
$$[(neg P land A) land (neg P land B)) lor (neg P land C) land (neg P land D))] lor [(R land A) land (R land B))lor (R land C) land (R land D))]$$
That is, you acted as if that last step was a Distribution ... but it isn't! ... because you end up distributing a $land$ over a $land$. Distribution is $land$ over $lor$ or vice versa.
So, you should have just stopped 1 line earlier, and thus have ended up with
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]$$
but since $land$ is Associative you can drop some parantheses and thus get:
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
and since $lor$ is Associative you can drop those square brackets as well:
$$(neg P land A land B)lor(neg 𝑃 land C land D)lor (R land A land B) lor (R land C land D)$$
answered Feb 2 at 20:34
Bram28Bram28
64.5k44793
$endgroup$
$begingroup$
Your edit clarifies it for me. Thank you!
$endgroup$
– tree_traversal
Feb 2 at 20:55
1
$begingroup$
@tree_traversal OK ... added just a little more at the end ... glad I could help!
$endgroup$
– Bram28
Feb 2 at 20:55
add a comment |
$begingroup$
Case 1 is right. Case 2 is not even a formula since some parentheses are missing: in $(lnot P lor A) land (lnot P land B) lor (lnot P lor C) land (lnot P lor D)$ it is not clear what is the principal connective.
Let us see why case 1 is correct. Essentially, you use repeatedly the following distributive law (together with associativity and commutativity):
begin{align}
A land (B lor C) &equiv (A land B) lor (A land C)
end{align}
By applying step-by-step this equivalence rule, you get:
begin{align}
&(neg P lor R) color{red}{land} big((Aland B) color{red}{lor} (Cland D)big) \
text{(distributivity) } equiv & ((neg P lor R) land (A land B)) lor ((neg P lor R) land (C land D)) \
text{(commutativity) } equiv & ((A land B) color{red}{land} (neg P color{red}{lor} R)) lor ((C land D) color{red}{land} (neg P color{red}{lor} R)) \
text{(distributivity) } equiv & big(((A land B) land neg P) lor ((A land B) land R)big) lor big(((C land D) land lnot P) lor ((C land D) land R)big) \
text{(associativity) } equiv & big((A land B land neg P) lor (A land B land R)big) lor big((C land D land lnot P) lor (C land D land R)big)
end{align}
answered Feb 2 at 21:08
TaroccoesbroccoTaroccoesbrocco
5,87471840
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Case 1 is correct.
What you end up with in case 2 is ungrammatical ... you'll need to add some parentheses. I assume you meant:
$$[color{red}((neg P color{red}land A) land (neg P land B)color{red})lorcolor{red}((neg 𝑃 land C)∧(neg 𝑃 land Dcolor{red})color{red})]lor [color{red}((R land A)land (R land B)) lor color{red}((R land C) land (R land D)color{red})]$$
But note that that is equivalent to:
$$[((neg P land A) land (neg P land B))lor((neg 𝑃 land C)land(neg 𝑃 land D))]lor [((R land A)land (R land B)) lor ((R land C) land (R land D))] overset{Association (drop parentheses in generalized conjunctions)}Leftrightarrow$$
$$[(neg P land A land neg P land B)lor(neg 𝑃 land Cland neg 𝑃 land D)]lor [(R land Aland R land B) lor (R land C land R land D)] overset{Commutation}Leftrightarrow$$
$$[(neg P land neg P land A land B)lor(neg 𝑃 land neg 𝑃 land C land D)]lor [(R land R land A land B) lor (R land R land C land D)] overset{Idempotence}Leftrightarrow$$
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
... which is exactly what you got from case 1
In other words, had you used proper parentheses, both results would have been correct!
That said, I have a feeling you did the following to get to case 2:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg P land ((Aland B)lor (Cland D))] lor [(R land ((Aland B)lor (Cland D))]=$$
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]=$$
$$[(neg P land A) land (neg P land B)) lor (neg P land C) land (neg P land D))] lor [(R land A) land (R land B))lor (R land C) land (R land D))]$$
That is, you acted as if that last step was a Distribution ... but it isn't! ... because you end up distributing a $land$ over a $land$. Distribution is $land$ over $lor$ or vice versa.
So, you should have just stopped 1 line earlier, and thus have ended up with
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]$$
but since $land$ is Associative you can drop some parantheses and thus get:
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
and since $lor$ is Associative you can drop those square brackets as well:
$$(neg P land A land B)lor(neg 𝑃 land C land D)lor (R land A land B) lor (R land C land D)$$
answered Feb 2 at 20:34
Bram28Bram28
64.5k44793
$endgroup$
$begingroup$
Your edit clarifies it for me. Thank you!
$endgroup$
– tree_traversal
Feb 2 at 20:55
1
$begingroup$
@tree_traversal OK ... added just a little more at the end ... glad I could help!
$endgroup$
– Bram28
Feb 2 at 20:55
add a comment |
$begingroup$
Case 1 is correct.
What you end up with in case 2 is ungrammatical ... you'll need to add some parentheses. I assume you meant:
$$[color{red}((neg P color{red}land A) land (neg P land B)color{red})lorcolor{red}((neg 𝑃 land C)∧(neg 𝑃 land Dcolor{red})color{red})]lor [color{red}((R land A)land (R land B)) lor color{red}((R land C) land (R land D)color{red})]$$
But note that that is equivalent to:
$$[((neg P land A) land (neg P land B))lor((neg 𝑃 land C)land(neg 𝑃 land D))]lor [((R land A)land (R land B)) lor ((R land C) land (R land D))] overset{Association (drop parentheses in generalized conjunctions)}Leftrightarrow$$
$$[(neg P land A land neg P land B)lor(neg 𝑃 land Cland neg 𝑃 land D)]lor [(R land Aland R land B) lor (R land C land R land D)] overset{Commutation}Leftrightarrow$$
$$[(neg P land neg P land A land B)lor(neg 𝑃 land neg 𝑃 land C land D)]lor [(R land R land A land B) lor (R land R land C land D)] overset{Idempotence}Leftrightarrow$$
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
... which is exactly what you got from case 1
In other words, had you used proper parentheses, both results would have been correct!
That said, I have a feeling you did the following to get to case 2:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg P land ((Aland B)lor (Cland D))] lor [(R land ((Aland B)lor (Cland D))]=$$
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]=$$
$$[(neg P land A) land (neg P land B)) lor (neg P land C) land (neg P land D))] lor [(R land A) land (R land B))lor (R land C) land (R land D))]$$
That is, you acted as if that last step was a Distribution ... but it isn't! ... because you end up distributing a $land$ over a $land$. Distribution is $land$ over $lor$ or vice versa.
So, you should have just stopped 1 line earlier, and thus have ended up with
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]$$
but since $land$ is Associative you can drop some parantheses and thus get:
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
and since $lor$ is Associative you can drop those square brackets as well:
$$(neg P land A land B)lor(neg 𝑃 land C land D)lor (R land A land B) lor (R land C land D)$$
answered Feb 2 at 20:34
Bram28Bram28
64.5k44793
$endgroup$
$begingroup$
Your edit clarifies it for me. Thank you!
$endgroup$
– tree_traversal
Feb 2 at 20:55
1
$begingroup$
@tree_traversal OK ... added just a little more at the end ... glad I could help!
$endgroup$
– Bram28
Feb 2 at 20:55
add a comment |
$begingroup$
Case 1 is correct.
What you end up with in case 2 is ungrammatical ... you'll need to add some parentheses. I assume you meant:
$$[color{red}((neg P color{red}land A) land (neg P land B)color{red})lorcolor{red}((neg 𝑃 land C)∧(neg 𝑃 land Dcolor{red})color{red})]lor [color{red}((R land A)land (R land B)) lor color{red}((R land C) land (R land D)color{red})]$$
But note that that is equivalent to:
$$[((neg P land A) land (neg P land B))lor((neg 𝑃 land C)land(neg 𝑃 land D))]lor [((R land A)land (R land B)) lor ((R land C) land (R land D))] overset{Association (drop parentheses in generalized conjunctions)}Leftrightarrow$$
$$[(neg P land A land neg P land B)lor(neg 𝑃 land Cland neg 𝑃 land D)]lor [(R land Aland R land B) lor (R land C land R land D)] overset{Commutation}Leftrightarrow$$
$$[(neg P land neg P land A land B)lor(neg 𝑃 land neg 𝑃 land C land D)]lor [(R land R land A land B) lor (R land R land C land D)] overset{Idempotence}Leftrightarrow$$
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
... which is exactly what you got from case 1
In other words, had you used proper parentheses, both results would have been correct!
That said, I have a feeling you did the following to get to case 2:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg P land ((Aland B)lor (Cland D))] lor [(R land ((Aland B)lor (Cland D))]=$$
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]=$$
$$[(neg P land A) land (neg P land B)) lor (neg P land C) land (neg P land D))] lor [(R land A) land (R land B))lor (R land C) land (R land D))]$$
That is, you acted as if that last step was a Distribution ... but it isn't! ... because you end up distributing a $land$ over a $land$. Distribution is $land$ over $lor$ or vice versa.
So, you should have just stopped 1 line earlier, and thus have ended up with
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]$$
but since $land$ is Associative you can drop some parantheses and thus get:
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
and since $lor$ is Associative you can drop those square brackets as well:
$$(neg P land A land B)lor(neg 𝑃 land C land D)lor (R land A land B) lor (R land C land D)$$
answered Feb 2 at 20:34
Bram28Bram28
64.5k44793
$endgroup$
Case 1 is correct.
What you end up with in case 2 is ungrammatical ... you'll need to add some parentheses. I assume you meant:
$$[color{red}((neg P color{red}land A) land (neg P land B)color{red})lorcolor{red}((neg 𝑃 land C)∧(neg 𝑃 land Dcolor{red})color{red})]lor [color{red}((R land A)land (R land B)) lor color{red}((R land C) land (R land D)color{red})]$$
But note that that is equivalent to:
$$[((neg P land A) land (neg P land B))lor((neg 𝑃 land C)land(neg 𝑃 land D))]lor [((R land A)land (R land B)) lor ((R land C) land (R land D))] overset{Association (drop parentheses in generalized conjunctions)}Leftrightarrow$$
$$[(neg P land A land neg P land B)lor(neg 𝑃 land Cland neg 𝑃 land D)]lor [(R land Aland R land B) lor (R land C land R land D)] overset{Commutation}Leftrightarrow$$
$$[(neg P land neg P land A land B)lor(neg 𝑃 land neg 𝑃 land C land D)]lor [(R land R land A land B) lor (R land R land C land D)] overset{Idempotence}Leftrightarrow$$
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
... which is exactly what you got from case 1
In other words, had you used proper parentheses, both results would have been correct!
That said, I have a feeling you did the following to get to case 2:
$$(neg P lor R) land [(Aland B)lor (Cland D)]=$$
$$[(neg P land ((Aland B)lor (Cland D))] lor [(R land ((Aland B)lor (Cland D))]=$$
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]=$$
$$[(neg P land A) land (neg P land B)) lor (neg P land C) land (neg P land D))] lor [(R land A) land (R land B))lor (R land C) land (R land D))]$$
That is, you acted as if that last step was a Distribution ... but it isn't! ... because you end up distributing a $land$ over a $land$. Distribution is $land$ over $lor$ or vice versa.
So, you should have just stopped 1 line earlier, and thus have ended up with
$$[(neg P land (Aland B)) lor (neg P land (Cland D))] lor [(R land (Aland B))lor (R land (Cland D))]$$
but since $land$ is Associative you can drop some parantheses and thus get:
$$[(neg P land A land B)lor(neg 𝑃 land C land D)]lor [(R land A land B) lor (R land C land D)]$$
and since $lor$ is Associative you can drop those square brackets as well:
$$(neg P land A land B)lor(neg 𝑃 land C land D)lor (R land A land B) lor (R land C land D)$$
answered Feb 2 at 20:34
Bram28Bram28
64.5k44793
edited Feb 2 at 20:55
answered Feb 2 at 20:34
Bram28Bram28
64.5k44793
answered Feb 2 at 20:34
Bram28Bram28
64.5k44793
answered Feb 2 at 20:34
Bram28Bram28
64.5k44793
64.5k44793
$begingroup$
Your edit clarifies it for me. Thank you!
$endgroup$
– tree_traversal
Feb 2 at 20:55
1
$begingroup$
@tree_traversal OK ... added just a little more at the end ... glad I could help!
$endgroup$
– Bram28
Feb 2 at 20:55
add a comment |
$begingroup$
Your edit clarifies it for me. Thank you!
$endgroup$
– tree_traversal
Feb 2 at 20:55
1
$begingroup$
@tree_traversal OK ... added just a little more at the end ... glad I could help!
$endgroup$
– Bram28
Feb 2 at 20:55
$begingroup$
Your edit clarifies it for me. Thank you!
$endgroup$
– tree_traversal
Feb 2 at 20:55
$begingroup$
Your edit clarifies it for me. Thank you!
$endgroup$
– tree_traversal
Feb 2 at 20:55
1
1
$begingroup$
@tree_traversal OK ... added just a little more at the end ... glad I could help!
$endgroup$
– Bram28
Feb 2 at 20:55
$begingroup$
@tree_traversal OK ... added just a little more at the end ... glad I could help!
$endgroup$
– Bram28
Feb 2 at 20:55
add a comment |
$begingroup$
Case 1 is right. Case 2 is not even a formula since some parentheses are missing: in $(lnot P lor A) land (lnot P land B) lor (lnot P lor C) land (lnot P lor D)$ it is not clear what is the principal connective.
Let us see why case 1 is correct. Essentially, you use repeatedly the following distributive law (together with associativity and commutativity):
begin{align}
A land (B lor C) &equiv (A land B) lor (A land C)
end{align}
By applying step-by-step this equivalence rule, you get:
begin{align}
&(neg P lor R) color{red}{land} big((Aland B) color{red}{lor} (Cland D)big) \
text{(distributivity) } equiv & ((neg P lor R) land (A land B)) lor ((neg P lor R) land (C land D)) \
text{(commutativity) } equiv & ((A land B) color{red}{land} (neg P color{red}{lor} R)) lor ((C land D) color{red}{land} (neg P color{red}{lor} R)) \
text{(distributivity) } equiv & big(((A land B) land neg P) lor ((A land B) land R)big) lor big(((C land D) land lnot P) lor ((C land D) land R)big) \
text{(associativity) } equiv & big((A land B land neg P) lor (A land B land R)big) lor big((C land D land lnot P) lor (C land D land R)big)
end{align}
answered Feb 2 at 21:08
TaroccoesbroccoTaroccoesbrocco
5,87471840
$endgroup$
add a comment |
$begingroup$
Case 1 is right. Case 2 is not even a formula since some parentheses are missing: in $(lnot P lor A) land (lnot P land B) lor (lnot P lor C) land (lnot P lor D)$ it is not clear what is the principal connective.
Let us see why case 1 is correct. Essentially, you use repeatedly the following distributive law (together with associativity and commutativity):
begin{align}
A land (B lor C) &equiv (A land B) lor (A land C)
end{align}
By applying step-by-step this equivalence rule, you get:
begin{align}
&(neg P lor R) color{red}{land} big((Aland B) color{red}{lor} (Cland D)big) \
text{(distributivity) } equiv & ((neg P lor R) land (A land B)) lor ((neg P lor R) land (C land D)) \
text{(commutativity) } equiv & ((A land B) color{red}{land} (neg P color{red}{lor} R)) lor ((C land D) color{red}{land} (neg P color{red}{lor} R)) \
text{(distributivity) } equiv & big(((A land B) land neg P) lor ((A land B) land R)big) lor big(((C land D) land lnot P) lor ((C land D) land R)big) \
text{(associativity) } equiv & big((A land B land neg P) lor (A land B land R)big) lor big((C land D land lnot P) lor (C land D land R)big)
end{align}
answered Feb 2 at 21:08
TaroccoesbroccoTaroccoesbrocco
5,87471840
$endgroup$
add a comment |
$begingroup$
Case 1 is right. Case 2 is not even a formula since some parentheses are missing: in $(lnot P lor A) land (lnot P land B) lor (lnot P lor C) land (lnot P lor D)$ it is not clear what is the principal connective.
Let us see why case 1 is correct. Essentially, you use repeatedly the following distributive law (together with associativity and commutativity):
begin{align}
A land (B lor C) &equiv (A land B) lor (A land C)
end{align}
By applying step-by-step this equivalence rule, you get:
begin{align}
&(neg P lor R) color{red}{land} big((Aland B) color{red}{lor} (Cland D)big) \
text{(distributivity) } equiv & ((neg P lor R) land (A land B)) lor ((neg P lor R) land (C land D)) \
text{(commutativity) } equiv & ((A land B) color{red}{land} (neg P color{red}{lor} R)) lor ((C land D) color{red}{land} (neg P color{red}{lor} R)) \
text{(distributivity) } equiv & big(((A land B) land neg P) lor ((A land B) land R)big) lor big(((C land D) land lnot P) lor ((C land D) land R)big) \
text{(associativity) } equiv & big((A land B land neg P) lor (A land B land R)big) lor big((C land D land lnot P) lor (C land D land R)big)
end{align}
answered Feb 2 at 21:08
TaroccoesbroccoTaroccoesbrocco
5,87471840
$endgroup$
Case 1 is right. Case 2 is not even a formula since some parentheses are missing: in $(lnot P lor A) land (lnot P land B) lor (lnot P lor C) land (lnot P lor D)$ it is not clear what is the principal connective.
Let us see why case 1 is correct. Essentially, you use repeatedly the following distributive law (together with associativity and commutativity):
begin{align}
A land (B lor C) &equiv (A land B) lor (A land C)
end{align}
By applying step-by-step this equivalence rule, you get:
begin{align}
&(neg P lor R) color{red}{land} big((Aland B) color{red}{lor} (Cland D)big) \
text{(distributivity) } equiv & ((neg P lor R) land (A land B)) lor ((neg P lor R) land (C land D)) \
text{(commutativity) } equiv & ((A land B) color{red}{land} (neg P color{red}{lor} R)) lor ((C land D) color{red}{land} (neg P color{red}{lor} R)) \
text{(distributivity) } equiv & big(((A land B) land neg P) lor ((A land B) land R)big) lor big(((C land D) land lnot P) lor ((C land D) land R)big) \
text{(associativity) } equiv & big((A land B land neg P) lor (A land B land R)big) lor big((C land D land lnot P) lor (C land D land R)big)
end{align}
answered Feb 2 at 21:08
TaroccoesbroccoTaroccoesbrocco
5,87471840
edited Feb 3 at 8:20
answered Feb 2 at 21:08
TaroccoesbroccoTaroccoesbrocco
5,87471840
answered Feb 2 at 21:08
TaroccoesbroccoTaroccoesbrocco
5,87471840
answered Feb 2 at 21:08
TaroccoesbroccoTaroccoesbrocco
5,87471840
5,87471840
add a comment |
add a comment |
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