Distributivity of Conjunction
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I am having trouble with understanding how distributivity of conjunction applies in a problem where I must prove:
$$ neg (a wedge neg b vee neg c) = neg (a vee neg c) vee b wedge c$$
My proof so far goes
$$ neg (a wedge neg b vee neg c)$$
$$ = neg a vee neg (neg b vee neg c)$$
$$ = neg a vee neg neg b wedge neg neg c$$
$$ = neg a vee b wedge c$$
$$ = (neg a wedge c) vee (b wedge c)$$
$$ = neg(a vee neg c) vee b wedge c$$
I am not sure if the step from the 4th to 5th line is correct. The distributivity property as I understand it is $ (x wedge z) vee (y wedge z) = z wedge (x vee y)$. I feel like it shouldn't apply in this case because the situation I have in line 4 is more like $ (z wedge x) vee y$. Does the presence of the parentheses prevent me from using the distributive property or do they go away after I distribute the negative sign?
logic propositional-calculus
edited Sep 11 at 7:23
Filippo De Bortoli
1,047521
asked Sep 11 at 3:54
mattzhu
865
add a comment |
up vote
0
down vote
favorite
I am having trouble with understanding how distributivity of conjunction applies in a problem where I must prove:
$$ neg (a wedge neg b vee neg c) = neg (a vee neg c) vee b wedge c$$
My proof so far goes
$$ neg (a wedge neg b vee neg c)$$
$$ = neg a vee neg (neg b vee neg c)$$
$$ = neg a vee neg neg b wedge neg neg c$$
$$ = neg a vee b wedge c$$
$$ = (neg a wedge c) vee (b wedge c)$$
$$ = neg(a vee neg c) vee b wedge c$$
I am not sure if the step from the 4th to 5th line is correct. The distributivity property as I understand it is $ (x wedge z) vee (y wedge z) = z wedge (x vee y)$. I feel like it shouldn't apply in this case because the situation I have in line 4 is more like $ (z wedge x) vee y$. Does the presence of the parentheses prevent me from using the distributive property or do they go away after I distribute the negative sign?
logic propositional-calculus
edited Sep 11 at 7:23
Filippo De Bortoli
1,047521
asked Sep 11 at 3:54
mattzhu
865
1
Your formula $a land lnot b lor lnot c$ is ambiguous: does it stand for $(a land lnot b) lor lnot c$ or $a land (lnot b lor lnot c)$? Analguous problem with $lnot(a lor lnot c) lor b land c$.
– Taroccoesbrocco
Sep 11 at 5:50
There was a time when conjunction had higher operational precedence than disjunction. That convention has been depreciated for so long that it is confusing to read. Please bracket your conjunctions.
– Graham Kemp
Sep 11 at 6:33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am having trouble with understanding how distributivity of conjunction applies in a problem where I must prove:
$$ neg (a wedge neg b vee neg c) = neg (a vee neg c) vee b wedge c$$
My proof so far goes
$$ neg (a wedge neg b vee neg c)$$
$$ = neg a vee neg (neg b vee neg c)$$
$$ = neg a vee neg neg b wedge neg neg c$$
$$ = neg a vee b wedge c$$
$$ = (neg a wedge c) vee (b wedge c)$$
$$ = neg(a vee neg c) vee b wedge c$$
I am not sure if the step from the 4th to 5th line is correct. The distributivity property as I understand it is $ (x wedge z) vee (y wedge z) = z wedge (x vee y)$. I feel like it shouldn't apply in this case because the situation I have in line 4 is more like $ (z wedge x) vee y$. Does the presence of the parentheses prevent me from using the distributive property or do they go away after I distribute the negative sign?
logic propositional-calculus
edited Sep 11 at 7:23
Filippo De Bortoli
1,047521
asked Sep 11 at 3:54
mattzhu
865
I am having trouble with understanding how distributivity of conjunction applies in a problem where I must prove:
$$ neg (a wedge neg b vee neg c) = neg (a vee neg c) vee b wedge c$$
My proof so far goes
$$ neg (a wedge neg b vee neg c)$$
$$ = neg a vee neg (neg b vee neg c)$$
$$ = neg a vee neg neg b wedge neg neg c$$
$$ = neg a vee b wedge c$$
$$ = (neg a wedge c) vee (b wedge c)$$
$$ = neg(a vee neg c) vee b wedge c$$
I am not sure if the step from the 4th to 5th line is correct. The distributivity property as I understand it is $ (x wedge z) vee (y wedge z) = z wedge (x vee y)$. I feel like it shouldn't apply in this case because the situation I have in line 4 is more like $ (z wedge x) vee y$. Does the presence of the parentheses prevent me from using the distributive property or do they go away after I distribute the negative sign?
logic propositional-calculus
logic propositional-calculus
edited Sep 11 at 7:23
Filippo De Bortoli
1,047521
asked Sep 11 at 3:54
mattzhu
865
edited Sep 11 at 7:23
Filippo De Bortoli
1,047521
asked Sep 11 at 3:54
mattzhu
865
edited Sep 11 at 7:23
Filippo De Bortoli
1,047521
edited Sep 11 at 7:23
Filippo De Bortoli
1,047521
edited Sep 11 at 7:23
Filippo De Bortoli
1,047521
1,047521
asked Sep 11 at 3:54
mattzhu
865
asked Sep 11 at 3:54
mattzhu
865
asked Sep 11 at 3:54
mattzhu
865
865
1
Your formula $a land lnot b lor lnot c$ is ambiguous: does it stand for $(a land lnot b) lor lnot c$ or $a land (lnot b lor lnot c)$? Analguous problem with $lnot(a lor lnot c) lor b land c$.
– Taroccoesbrocco
Sep 11 at 5:50
There was a time when conjunction had higher operational precedence than disjunction. That convention has been depreciated for so long that it is confusing to read. Please bracket your conjunctions.
– Graham Kemp
Sep 11 at 6:33
add a comment |
1
Your formula $a land lnot b lor lnot c$ is ambiguous: does it stand for $(a land lnot b) lor lnot c$ or $a land (lnot b lor lnot c)$? Analguous problem with $lnot(a lor lnot c) lor b land c$.
– Taroccoesbrocco
Sep 11 at 5:50
There was a time when conjunction had higher operational precedence than disjunction. That convention has been depreciated for so long that it is confusing to read. Please bracket your conjunctions.
– Graham Kemp
Sep 11 at 6:33
1
1
Your formula $a land lnot b lor lnot c$ is ambiguous: does it stand for $(a land lnot b) lor lnot c$ or $a land (lnot b lor lnot c)$? Analguous problem with $lnot(a lor lnot c) lor b land c$.
– Taroccoesbrocco
Sep 11 at 5:50
Your formula $a land lnot b lor lnot c$ is ambiguous: does it stand for $(a land lnot b) lor lnot c$ or $a land (lnot b lor lnot c)$? Analguous problem with $lnot(a lor lnot c) lor b land c$.
– Taroccoesbrocco
Sep 11 at 5:50
There was a time when conjunction had higher operational precedence than disjunction. That convention has been depreciated for so long that it is confusing to read. Please bracket your conjunctions.
– Graham Kemp
Sep 11 at 6:33
There was a time when conjunction had higher operational precedence than disjunction. That convention has been depreciated for so long that it is confusing to read. Please bracket your conjunctions.
– Graham Kemp
Sep 11 at 6:33
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
There are two distribution laws:
Conjunction distributes over disjunction: $xland (ylor z)=(xland y)lor(xland z)$
Disjunction distributes over conjunction: $xlor (yland z)=(xlor y)land(xlor z)$
Also recall that conjunction and disjunction are both commutative.
$xland (ylor z)=(ylor z)land x=(yland x)lor(zland x)=(xland y)lor(xland z)$
$xlor (yland z)=(yland z)lor x =$ et cetera
answered Sep 11 at 6:27
Graham Kemp
83.4k43378
Right. I would also add that in general $a lor (b land c) ne (a lor b) land c$; there seems to be a lot of ambiguity wrt precedence of operators in the OP's reasoning.
– Filippo De Bortoli
Sep 11 at 6:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There are two distribution laws:
Conjunction distributes over disjunction: $xland (ylor z)=(xland y)lor(xland z)$
Disjunction distributes over conjunction: $xlor (yland z)=(xlor y)land(xlor z)$
Also recall that conjunction and disjunction are both commutative.
$xland (ylor z)=(ylor z)land x=(yland x)lor(zland x)=(xland y)lor(xland z)$
$xlor (yland z)=(yland z)lor x =$ et cetera
answered Sep 11 at 6:27
Graham Kemp
83.4k43378
Right. I would also add that in general $a lor (b land c) ne (a lor b) land c$; there seems to be a lot of ambiguity wrt precedence of operators in the OP's reasoning.
– Filippo De Bortoli
Sep 11 at 6:34
add a comment |
up vote
1
down vote
There are two distribution laws:
Conjunction distributes over disjunction: $xland (ylor z)=(xland y)lor(xland z)$
Disjunction distributes over conjunction: $xlor (yland z)=(xlor y)land(xlor z)$
Also recall that conjunction and disjunction are both commutative.
$xland (ylor z)=(ylor z)land x=(yland x)lor(zland x)=(xland y)lor(xland z)$
$xlor (yland z)=(yland z)lor x =$ et cetera
answered Sep 11 at 6:27
Graham Kemp
83.4k43378
Right. I would also add that in general $a lor (b land c) ne (a lor b) land c$; there seems to be a lot of ambiguity wrt precedence of operators in the OP's reasoning.
– Filippo De Bortoli
Sep 11 at 6:34
add a comment |
up vote
1
down vote
up vote
1
down vote
There are two distribution laws:
Conjunction distributes over disjunction: $xland (ylor z)=(xland y)lor(xland z)$
Disjunction distributes over conjunction: $xlor (yland z)=(xlor y)land(xlor z)$
Also recall that conjunction and disjunction are both commutative.
$xland (ylor z)=(ylor z)land x=(yland x)lor(zland x)=(xland y)lor(xland z)$
$xlor (yland z)=(yland z)lor x =$ et cetera
answered Sep 11 at 6:27
Graham Kemp
83.4k43378
There are two distribution laws:
Conjunction distributes over disjunction: $xland (ylor z)=(xland y)lor(xland z)$
Disjunction distributes over conjunction: $xlor (yland z)=(xlor y)land(xlor z)$
Also recall that conjunction and disjunction are both commutative.
$xland (ylor z)=(ylor z)land x=(yland x)lor(zland x)=(xland y)lor(xland z)$
$xlor (yland z)=(yland z)lor x =$ et cetera
answered Sep 11 at 6:27
Graham Kemp
83.4k43378
answered Sep 11 at 6:27
Graham Kemp
83.4k43378
answered Sep 11 at 6:27
Graham Kemp
83.4k43378
answered Sep 11 at 6:27
Graham Kemp
83.4k43378
83.4k43378
Right. I would also add that in general $a lor (b land c) ne (a lor b) land c$; there seems to be a lot of ambiguity wrt precedence of operators in the OP's reasoning.
– Filippo De Bortoli
Sep 11 at 6:34
add a comment |
Right. I would also add that in general $a lor (b land c) ne (a lor b) land c$; there seems to be a lot of ambiguity wrt precedence of operators in the OP's reasoning.
– Filippo De Bortoli
Sep 11 at 6:34
Right. I would also add that in general $a lor (b land c) ne (a lor b) land c$; there seems to be a lot of ambiguity wrt precedence of operators in the OP's reasoning.
– Filippo De Bortoli
Sep 11 at 6:34
Right. I would also add that in general $a lor (b land c) ne (a lor b) land c$; there seems to be a lot of ambiguity wrt precedence of operators in the OP's reasoning.
– Filippo De Bortoli
Sep 11 at 6:34
add a comment |
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1
Your formula $a land lnot b lor lnot c$ is ambiguous: does it stand for $(a land lnot b) lor lnot c$ or $a land (lnot b lor lnot c)$? Analguous problem with $lnot(a lor lnot c) lor b land c$.
– Taroccoesbrocco
Sep 11 at 5:50
There was a time when conjunction had higher operational precedence than disjunction. That convention has been depreciated for so long that it is confusing to read. Please bracket your conjunctions.
– Graham Kemp
Sep 11 at 6:33