Mixing
Show that $(P to R) land (Q to R)$ is equivalent to $(P lor Q) to R$
$begingroup$
I have checked the answer, but do not understand one step.
$(P to R) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
which is equivalent to $(neg P land neg Q ) lor R$
I don't understand the second equivalent. How to arrive to this?
logic
asked Feb 2 at 1:50
JOHN JOHN
4589
$endgroup$
add a comment |
$begingroup$
I have checked the answer, but do not understand one step.
$(P to R) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
which is equivalent to $(neg P land neg Q ) lor R$
I don't understand the second equivalent. How to arrive to this?
logic
asked Feb 2 at 1:50
JOHN JOHN
4589
$endgroup$
2
$begingroup$
As @DerekElkins pointed out, what you're trying to prove is false. Your attempt suggests that the problem actually involved $(Pto R)land (Qto R)$, which is equivalent to $(Plor Q)to R$.
$endgroup$
– Andreas Blass
Feb 2 at 3:30
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@AndreasBlass Sorry. I have a typo
$endgroup$
– JOHN
Feb 2 at 6:25
add a comment |
$begingroup$
I have checked the answer, but do not understand one step.
$(P to R) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
which is equivalent to $(neg P land neg Q ) lor R$
I don't understand the second equivalent. How to arrive to this?
logic
asked Feb 2 at 1:50
JOHN JOHN
4589
$endgroup$
I have checked the answer, but do not understand one step.
$(P to R) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
which is equivalent to $(neg P land neg Q ) lor R$
I don't understand the second equivalent. How to arrive to this?
logic
logic
asked Feb 2 at 1:50
JOHN JOHN
4589
asked Feb 2 at 1:50
JOHN JOHN
4589
edited Feb 2 at 6:47
JOHN
asked Feb 2 at 1:50
JOHN JOHN
4589
asked Feb 2 at 1:50
JOHN JOHN
4589
asked Feb 2 at 1:50
JOHN JOHN
4589
4589
2
$begingroup$
As @DerekElkins pointed out, what you're trying to prove is false. Your attempt suggests that the problem actually involved $(Pto R)land (Qto R)$, which is equivalent to $(Plor Q)to R$.
$endgroup$
– Andreas Blass
Feb 2 at 3:30
$begingroup$
@AndreasBlass Sorry. I have a typo
$endgroup$
– JOHN
Feb 2 at 6:25
add a comment |
2
$begingroup$
As @DerekElkins pointed out, what you're trying to prove is false. Your attempt suggests that the problem actually involved $(Pto R)land (Qto R)$, which is equivalent to $(Plor Q)to R$.
$endgroup$
– Andreas Blass
Feb 2 at 3:30
$begingroup$
@AndreasBlass Sorry. I have a typo
$endgroup$
– JOHN
Feb 2 at 6:25
2
2
$begingroup$
As @DerekElkins pointed out, what you're trying to prove is false. Your attempt suggests that the problem actually involved $(Pto R)land (Qto R)$, which is equivalent to $(Plor Q)to R$.
$endgroup$
– Andreas Blass
Feb 2 at 3:30
$begingroup$
As @DerekElkins pointed out, what you're trying to prove is false. Your attempt suggests that the problem actually involved $(Pto R)land (Qto R)$, which is equivalent to $(Plor Q)to R$.
$endgroup$
– Andreas Blass
Feb 2 at 3:30
$begingroup$
@AndreasBlass Sorry. I have a typo
$endgroup$
– JOHN
Feb 2 at 6:25
$begingroup$
@AndreasBlass Sorry. I have a typo
$endgroup$
– JOHN
Feb 2 at 6:25
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
A different approach for the problem is to convert P,Q,R to Boolean variables and form a truth table where $0$ represents False and $1$ represents True.
begin{array}{|c|c|} hline
P & Q & R & Prightarrow Q & Qrightarrow R & (Prightarrow Q)land(Q rightarrow R) &Plor Q &(P lor Q)rightarrow R\ hline
0 & 0 & 0& 1& 1& 1& 0& 1 \ hline
0 & 0 & 1& 1& 1& 1& 0& 1 \ hline
0 & 1 & 0& 1& 0& 0& 1& 0 \ hline
0 & 1 & 1& 1& 1& 1& 1& 1 \ hline
1 & 0 & 0& 0& 1& 0& 1& 0 \ hline
1 & 0 & 1& 0& 1& 0& 1& 0 \ hline
1 & 1 & 0& 1& 0& 0& 1& 0 \ hline
1 & 1 & 1& 1& 1& 1& 1& 1 \ hline
end{array}
So both the statements are equivalent. Hope this helps...
Another approach is to replace $Prightarrow Q$ by $neg P vee Q$ and solve the expression
answered Feb 2 at 3:29
SNEHIL SANYALSNEHIL SANYAL
668110
$endgroup$
add a comment |
$begingroup$
$(P to Q) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
No, that is meant to be $(neg P lor Q) land (neg Q lor R)$ .
which is equivalent to $(neg P land neg Q ) lor R$
$$begin{align}&(neg P lor Q) land (neg Q lor R)
\=~&((lnot Plor Q)landlnot Q)lor((lnot Plor Q)land R)\=~& (lnot Plandlnot Q)lor((lnot Plor Q)land R)\=&~ ((lnot Plandlnot Q)lor(lnot Plor Q))land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)lor Rend{align}$$
answered Feb 2 at 5:42
Graham KempGraham Kemp
87.9k43578
$endgroup$
$begingroup$
I don't think the second line correct.
$endgroup$
– Doug Spoonwood
Feb 2 at 5:52
$begingroup$
Why the second line $(lnot Pland lnot Q)lor (Pland R)lor (Qlandlnot Q)lor(Rland R)\$ is so?
$endgroup$
– JOHN
Feb 2 at 6:24
$begingroup$
@GrahamKemp I am really sorry. I messed up the symbol in the question. Can you modify the answer again? It is close.
$endgroup$
– JOHN
Feb 2 at 6:49
add a comment |
$begingroup$
One way is to take each pair of expressions and try to reconcile them using known rules, e.g. De Morgan's laws.
The second approach is brute force, namely using truth tables. Not particularly elegant, but it works.
In your case, the unknown variables are P, Q and R. You take 2^3 = 8 combinations of true/false for each variable, and check all your formulas yield the same answer.
You may check this by hand. Example code for Mathematica
formulaA[p_, q_, r_] := (p~Implies~q)~And~(q~Implies~r);
formulaB[p_, q_, r_] := (Not[p]~And~r)~And~(Not[q]~Or~r);
formulaC[p_, q_, r_] := (Not[p]~And~Nor[q])~Or~r;
BooleanTable[{p, q, r} -> formulaA[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaB[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaC[p, q, r], {p, q, r}] // TableForm
It yields something like this. Did you write your first formula correct?
{p,q,r} -> result
{True,True,True}->True
{True,True,False}->False
{True,False,True}->True
{True,False,False}->False
{False,True,True}->True
{False,True,False}->False
{False,False,True}->True
{False,False,False}->True
answered Feb 2 at 2:27
Mikhail DMikhail D
36326
$endgroup$
add a comment |
$begingroup$
Work with truth values $1,,0$ instead of true, false, so $Plor Q=max{P,,Q}$. Then the claim is that $P,,Qle R$ is equivalent to $max{P,,Q}le R$, which is trivial.
answered Feb 2 at 10:04
J.G.J.G.
33.3k23252
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A different approach for the problem is to convert P,Q,R to Boolean variables and form a truth table where $0$ represents False and $1$ represents True.
begin{array}{|c|c|} hline
P & Q & R & Prightarrow Q & Qrightarrow R & (Prightarrow Q)land(Q rightarrow R) &Plor Q &(P lor Q)rightarrow R\ hline
0 & 0 & 0& 1& 1& 1& 0& 1 \ hline
0 & 0 & 1& 1& 1& 1& 0& 1 \ hline
0 & 1 & 0& 1& 0& 0& 1& 0 \ hline
0 & 1 & 1& 1& 1& 1& 1& 1 \ hline
1 & 0 & 0& 0& 1& 0& 1& 0 \ hline
1 & 0 & 1& 0& 1& 0& 1& 0 \ hline
1 & 1 & 0& 1& 0& 0& 1& 0 \ hline
1 & 1 & 1& 1& 1& 1& 1& 1 \ hline
end{array}
So both the statements are equivalent. Hope this helps...
Another approach is to replace $Prightarrow Q$ by $neg P vee Q$ and solve the expression
answered Feb 2 at 3:29
SNEHIL SANYALSNEHIL SANYAL
668110
$endgroup$
add a comment |
$begingroup$
A different approach for the problem is to convert P,Q,R to Boolean variables and form a truth table where $0$ represents False and $1$ represents True.
begin{array}{|c|c|} hline
P & Q & R & Prightarrow Q & Qrightarrow R & (Prightarrow Q)land(Q rightarrow R) &Plor Q &(P lor Q)rightarrow R\ hline
0 & 0 & 0& 1& 1& 1& 0& 1 \ hline
0 & 0 & 1& 1& 1& 1& 0& 1 \ hline
0 & 1 & 0& 1& 0& 0& 1& 0 \ hline
0 & 1 & 1& 1& 1& 1& 1& 1 \ hline
1 & 0 & 0& 0& 1& 0& 1& 0 \ hline
1 & 0 & 1& 0& 1& 0& 1& 0 \ hline
1 & 1 & 0& 1& 0& 0& 1& 0 \ hline
1 & 1 & 1& 1& 1& 1& 1& 1 \ hline
end{array}
So both the statements are equivalent. Hope this helps...
Another approach is to replace $Prightarrow Q$ by $neg P vee Q$ and solve the expression
answered Feb 2 at 3:29
SNEHIL SANYALSNEHIL SANYAL
668110
$endgroup$
add a comment |
$begingroup$
A different approach for the problem is to convert P,Q,R to Boolean variables and form a truth table where $0$ represents False and $1$ represents True.
begin{array}{|c|c|} hline
P & Q & R & Prightarrow Q & Qrightarrow R & (Prightarrow Q)land(Q rightarrow R) &Plor Q &(P lor Q)rightarrow R\ hline
0 & 0 & 0& 1& 1& 1& 0& 1 \ hline
0 & 0 & 1& 1& 1& 1& 0& 1 \ hline
0 & 1 & 0& 1& 0& 0& 1& 0 \ hline
0 & 1 & 1& 1& 1& 1& 1& 1 \ hline
1 & 0 & 0& 0& 1& 0& 1& 0 \ hline
1 & 0 & 1& 0& 1& 0& 1& 0 \ hline
1 & 1 & 0& 1& 0& 0& 1& 0 \ hline
1 & 1 & 1& 1& 1& 1& 1& 1 \ hline
end{array}
So both the statements are equivalent. Hope this helps...
Another approach is to replace $Prightarrow Q$ by $neg P vee Q$ and solve the expression
answered Feb 2 at 3:29
SNEHIL SANYALSNEHIL SANYAL
668110
$endgroup$
A different approach for the problem is to convert P,Q,R to Boolean variables and form a truth table where $0$ represents False and $1$ represents True.
begin{array}{|c|c|} hline
P & Q & R & Prightarrow Q & Qrightarrow R & (Prightarrow Q)land(Q rightarrow R) &Plor Q &(P lor Q)rightarrow R\ hline
0 & 0 & 0& 1& 1& 1& 0& 1 \ hline
0 & 0 & 1& 1& 1& 1& 0& 1 \ hline
0 & 1 & 0& 1& 0& 0& 1& 0 \ hline
0 & 1 & 1& 1& 1& 1& 1& 1 \ hline
1 & 0 & 0& 0& 1& 0& 1& 0 \ hline
1 & 0 & 1& 0& 1& 0& 1& 0 \ hline
1 & 1 & 0& 1& 0& 0& 1& 0 \ hline
1 & 1 & 1& 1& 1& 1& 1& 1 \ hline
end{array}
So both the statements are equivalent. Hope this helps...
Another approach is to replace $Prightarrow Q$ by $neg P vee Q$ and solve the expression
answered Feb 2 at 3:29
SNEHIL SANYALSNEHIL SANYAL
668110
answered Feb 2 at 3:29
SNEHIL SANYALSNEHIL SANYAL
668110
answered Feb 2 at 3:29
SNEHIL SANYALSNEHIL SANYAL
668110
answered Feb 2 at 3:29
SNEHIL SANYALSNEHIL SANYAL
668110
668110
add a comment |
add a comment |
$begingroup$
$(P to Q) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
No, that is meant to be $(neg P lor Q) land (neg Q lor R)$ .
which is equivalent to $(neg P land neg Q ) lor R$
$$begin{align}&(neg P lor Q) land (neg Q lor R)
\=~&((lnot Plor Q)landlnot Q)lor((lnot Plor Q)land R)\=~& (lnot Plandlnot Q)lor((lnot Plor Q)land R)\=&~ ((lnot Plandlnot Q)lor(lnot Plor Q))land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)lor Rend{align}$$
answered Feb 2 at 5:42
Graham KempGraham Kemp
87.9k43578
$endgroup$
$begingroup$
I don't think the second line correct.
$endgroup$
– Doug Spoonwood
Feb 2 at 5:52
$begingroup$
Why the second line $(lnot Pland lnot Q)lor (Pland R)lor (Qlandlnot Q)lor(Rland R)\$ is so?
$endgroup$
– JOHN
Feb 2 at 6:24
$begingroup$
@GrahamKemp I am really sorry. I messed up the symbol in the question. Can you modify the answer again? It is close.
$endgroup$
– JOHN
Feb 2 at 6:49
add a comment |
$begingroup$
$(P to Q) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
No, that is meant to be $(neg P lor Q) land (neg Q lor R)$ .
which is equivalent to $(neg P land neg Q ) lor R$
$$begin{align}&(neg P lor Q) land (neg Q lor R)
\=~&((lnot Plor Q)landlnot Q)lor((lnot Plor Q)land R)\=~& (lnot Plandlnot Q)lor((lnot Plor Q)land R)\=&~ ((lnot Plandlnot Q)lor(lnot Plor Q))land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)lor Rend{align}$$
answered Feb 2 at 5:42
Graham KempGraham Kemp
87.9k43578
$endgroup$
$begingroup$
I don't think the second line correct.
$endgroup$
– Doug Spoonwood
Feb 2 at 5:52
$begingroup$
Why the second line $(lnot Pland lnot Q)lor (Pland R)lor (Qlandlnot Q)lor(Rland R)\$ is so?
$endgroup$
– JOHN
Feb 2 at 6:24
$begingroup$
@GrahamKemp I am really sorry. I messed up the symbol in the question. Can you modify the answer again? It is close.
$endgroup$
– JOHN
Feb 2 at 6:49
add a comment |
$begingroup$
$(P to Q) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
No, that is meant to be $(neg P lor Q) land (neg Q lor R)$ .
which is equivalent to $(neg P land neg Q ) lor R$
$$begin{align}&(neg P lor Q) land (neg Q lor R)
\=~&((lnot Plor Q)landlnot Q)lor((lnot Plor Q)land R)\=~& (lnot Plandlnot Q)lor((lnot Plor Q)land R)\=&~ ((lnot Plandlnot Q)lor(lnot Plor Q))land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)lor Rend{align}$$
answered Feb 2 at 5:42
Graham KempGraham Kemp
87.9k43578
$endgroup$
$(P to Q) land (Q to R)$ is equivalent to $(neg P lor R) land (neg Q lor R)$
No, that is meant to be $(neg P lor Q) land (neg Q lor R)$ .
which is equivalent to $(neg P land neg Q ) lor R$
$$begin{align}&(neg P lor Q) land (neg Q lor R)
\=~&((lnot Plor Q)landlnot Q)lor((lnot Plor Q)land R)\=~& (lnot Plandlnot Q)lor((lnot Plor Q)land R)\=&~ ((lnot Plandlnot Q)lor(lnot Plor Q))land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)land((lnot Plandlnot Q)lor R)\=&~ (lnot Plandlnot Q)lor Rend{align}$$
answered Feb 2 at 5:42
Graham KempGraham Kemp
87.9k43578
edited Feb 2 at 7:03
answered Feb 2 at 5:42
Graham KempGraham Kemp
87.9k43578
answered Feb 2 at 5:42
Graham KempGraham Kemp
87.9k43578
answered Feb 2 at 5:42
Graham KempGraham Kemp
87.9k43578
87.9k43578
$begingroup$
I don't think the second line correct.
$endgroup$
– Doug Spoonwood
Feb 2 at 5:52
$begingroup$
Why the second line $(lnot Pland lnot Q)lor (Pland R)lor (Qlandlnot Q)lor(Rland R)\$ is so?
$endgroup$
– JOHN
Feb 2 at 6:24
$begingroup$
@GrahamKemp I am really sorry. I messed up the symbol in the question. Can you modify the answer again? It is close.
$endgroup$
– JOHN
Feb 2 at 6:49
add a comment |
$begingroup$
I don't think the second line correct.
$endgroup$
– Doug Spoonwood
Feb 2 at 5:52
$begingroup$
Why the second line $(lnot Pland lnot Q)lor (Pland R)lor (Qlandlnot Q)lor(Rland R)\$ is so?
$endgroup$
– JOHN
Feb 2 at 6:24
$begingroup$
@GrahamKemp I am really sorry. I messed up the symbol in the question. Can you modify the answer again? It is close.
$endgroup$
– JOHN
Feb 2 at 6:49
$begingroup$
I don't think the second line correct.
$endgroup$
– Doug Spoonwood
Feb 2 at 5:52
$begingroup$
I don't think the second line correct.
$endgroup$
– Doug Spoonwood
Feb 2 at 5:52
$begingroup$
Why the second line $(lnot Pland lnot Q)lor (Pland R)lor (Qlandlnot Q)lor(Rland R)\$ is so?
$endgroup$
– JOHN
Feb 2 at 6:24
$begingroup$
Why the second line $(lnot Pland lnot Q)lor (Pland R)lor (Qlandlnot Q)lor(Rland R)\$ is so?
$endgroup$
– JOHN
Feb 2 at 6:24
$begingroup$
@GrahamKemp I am really sorry. I messed up the symbol in the question. Can you modify the answer again? It is close.
$endgroup$
– JOHN
Feb 2 at 6:49
$begingroup$
@GrahamKemp I am really sorry. I messed up the symbol in the question. Can you modify the answer again? It is close.
$endgroup$
– JOHN
Feb 2 at 6:49
add a comment |
$begingroup$
One way is to take each pair of expressions and try to reconcile them using known rules, e.g. De Morgan's laws.
The second approach is brute force, namely using truth tables. Not particularly elegant, but it works.
In your case, the unknown variables are P, Q and R. You take 2^3 = 8 combinations of true/false for each variable, and check all your formulas yield the same answer.
You may check this by hand. Example code for Mathematica
formulaA[p_, q_, r_] := (p~Implies~q)~And~(q~Implies~r);
formulaB[p_, q_, r_] := (Not[p]~And~r)~And~(Not[q]~Or~r);
formulaC[p_, q_, r_] := (Not[p]~And~Nor[q])~Or~r;
BooleanTable[{p, q, r} -> formulaA[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaB[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaC[p, q, r], {p, q, r}] // TableForm
It yields something like this. Did you write your first formula correct?
{p,q,r} -> result
{True,True,True}->True
{True,True,False}->False
{True,False,True}->True
{True,False,False}->False
{False,True,True}->True
{False,True,False}->False
{False,False,True}->True
{False,False,False}->True
answered Feb 2 at 2:27
Mikhail DMikhail D
36326
$endgroup$
add a comment |
$begingroup$
One way is to take each pair of expressions and try to reconcile them using known rules, e.g. De Morgan's laws.
The second approach is brute force, namely using truth tables. Not particularly elegant, but it works.
In your case, the unknown variables are P, Q and R. You take 2^3 = 8 combinations of true/false for each variable, and check all your formulas yield the same answer.
You may check this by hand. Example code for Mathematica
formulaA[p_, q_, r_] := (p~Implies~q)~And~(q~Implies~r);
formulaB[p_, q_, r_] := (Not[p]~And~r)~And~(Not[q]~Or~r);
formulaC[p_, q_, r_] := (Not[p]~And~Nor[q])~Or~r;
BooleanTable[{p, q, r} -> formulaA[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaB[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaC[p, q, r], {p, q, r}] // TableForm
It yields something like this. Did you write your first formula correct?
{p,q,r} -> result
{True,True,True}->True
{True,True,False}->False
{True,False,True}->True
{True,False,False}->False
{False,True,True}->True
{False,True,False}->False
{False,False,True}->True
{False,False,False}->True
answered Feb 2 at 2:27
Mikhail DMikhail D
36326
$endgroup$
add a comment |
$begingroup$
One way is to take each pair of expressions and try to reconcile them using known rules, e.g. De Morgan's laws.
The second approach is brute force, namely using truth tables. Not particularly elegant, but it works.
In your case, the unknown variables are P, Q and R. You take 2^3 = 8 combinations of true/false for each variable, and check all your formulas yield the same answer.
You may check this by hand. Example code for Mathematica
formulaA[p_, q_, r_] := (p~Implies~q)~And~(q~Implies~r);
formulaB[p_, q_, r_] := (Not[p]~And~r)~And~(Not[q]~Or~r);
formulaC[p_, q_, r_] := (Not[p]~And~Nor[q])~Or~r;
BooleanTable[{p, q, r} -> formulaA[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaB[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaC[p, q, r], {p, q, r}] // TableForm
It yields something like this. Did you write your first formula correct?
{p,q,r} -> result
{True,True,True}->True
{True,True,False}->False
{True,False,True}->True
{True,False,False}->False
{False,True,True}->True
{False,True,False}->False
{False,False,True}->True
{False,False,False}->True
answered Feb 2 at 2:27
Mikhail DMikhail D
36326
$endgroup$
One way is to take each pair of expressions and try to reconcile them using known rules, e.g. De Morgan's laws.
The second approach is brute force, namely using truth tables. Not particularly elegant, but it works.
In your case, the unknown variables are P, Q and R. You take 2^3 = 8 combinations of true/false for each variable, and check all your formulas yield the same answer.
You may check this by hand. Example code for Mathematica
formulaA[p_, q_, r_] := (p~Implies~q)~And~(q~Implies~r);
formulaB[p_, q_, r_] := (Not[p]~And~r)~And~(Not[q]~Or~r);
formulaC[p_, q_, r_] := (Not[p]~And~Nor[q])~Or~r;
BooleanTable[{p, q, r} -> formulaA[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaB[p, q, r], {p, q, r}] // TableForm
BooleanTable[{p, q, r} -> formulaC[p, q, r], {p, q, r}] // TableForm
It yields something like this. Did you write your first formula correct?
{p,q,r} -> result
{True,True,True}->True
{True,True,False}->False
{True,False,True}->True
{True,False,False}->False
{False,True,True}->True
{False,True,False}->False
{False,False,True}->True
{False,False,False}->True
answered Feb 2 at 2:27
Mikhail DMikhail D
36326
answered Feb 2 at 2:27
Mikhail DMikhail D
36326
answered Feb 2 at 2:27
Mikhail DMikhail D
36326
answered Feb 2 at 2:27
Mikhail DMikhail D
36326
36326
add a comment |
add a comment |
$begingroup$
Work with truth values $1,,0$ instead of true, false, so $Plor Q=max{P,,Q}$. Then the claim is that $P,,Qle R$ is equivalent to $max{P,,Q}le R$, which is trivial.
answered Feb 2 at 10:04
J.G.J.G.
33.3k23252
$endgroup$
add a comment |
$begingroup$
Work with truth values $1,,0$ instead of true, false, so $Plor Q=max{P,,Q}$. Then the claim is that $P,,Qle R$ is equivalent to $max{P,,Q}le R$, which is trivial.
answered Feb 2 at 10:04
J.G.J.G.
33.3k23252
$endgroup$
add a comment |
$begingroup$
Work with truth values $1,,0$ instead of true, false, so $Plor Q=max{P,,Q}$. Then the claim is that $P,,Qle R$ is equivalent to $max{P,,Q}le R$, which is trivial.
answered Feb 2 at 10:04
J.G.J.G.
33.3k23252
$endgroup$
Work with truth values $1,,0$ instead of true, false, so $Plor Q=max{P,,Q}$. Then the claim is that $P,,Qle R$ is equivalent to $max{P,,Q}le R$, which is trivial.
answered Feb 2 at 10:04
J.G.J.G.
33.3k23252
answered Feb 2 at 10:04
J.G.J.G.
33.3k23252
answered Feb 2 at 10:04
J.G.J.G.
33.3k23252
answered Feb 2 at 10:04
J.G.J.G.
33.3k23252
33.3k23252
add a comment |
add a comment |
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$begingroup$
As @DerekElkins pointed out, what you're trying to prove is false. Your attempt suggests that the problem actually involved $(Pto R)land (Qto R)$, which is equivalent to $(Plor Q)to R$.
$endgroup$
– Andreas Blass
Feb 2 at 3:30
$begingroup$
@AndreasBlass Sorry. I have a typo
$endgroup$
– JOHN
Feb 2 at 6:25