Show that the conditional statement is a tautology without using a truth table
I have been attempting to use identities to get to the answer but I am unable to get anywhere.
Here is the equation I am trying to prove tautological without using truth tables:
$[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$
And here is my attempt at using identities to work my way up. (a link to a page with a table of logical equivalence identities: http://en.wikipedia.org/wiki/Logical_equivalence)
$lnot [(p rightarrow q) land (q rightarrow r)] lor (p rightarrow r)$
$lnot [(lnot p lor q) land (q rightarrow r)] lor (p rightarrow r)$
$[lnot (lnot p lor q) land lnot (q rightarrow r)] lor (p rightarrow r)$
$[(pland lnot q) land (q land lnot r)] lor (p rightarrow r)$
I don't know where to go from there.
logic propositional-calculus
add a comment |
I have been attempting to use identities to get to the answer but I am unable to get anywhere.
Here is the equation I am trying to prove tautological without using truth tables:
$[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$
And here is my attempt at using identities to work my way up. (a link to a page with a table of logical equivalence identities: http://en.wikipedia.org/wiki/Logical_equivalence)
$lnot [(p rightarrow q) land (q rightarrow r)] lor (p rightarrow r)$
$lnot [(lnot p lor q) land (q rightarrow r)] lor (p rightarrow r)$
$[lnot (lnot p lor q) land lnot (q rightarrow r)] lor (p rightarrow r)$
$[(pland lnot q) land (q land lnot r)] lor (p rightarrow r)$
I don't know where to go from there.
logic propositional-calculus
In fourth step you did not apply the nagasion properly due to nagasion sign the AND operator will change into OR operator and then the whole question will solve very easily.
– user620344
Nov 27 '18 at 18:55
add a comment |
I have been attempting to use identities to get to the answer but I am unable to get anywhere.
Here is the equation I am trying to prove tautological without using truth tables:
$[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$
And here is my attempt at using identities to work my way up. (a link to a page with a table of logical equivalence identities: http://en.wikipedia.org/wiki/Logical_equivalence)
$lnot [(p rightarrow q) land (q rightarrow r)] lor (p rightarrow r)$
$lnot [(lnot p lor q) land (q rightarrow r)] lor (p rightarrow r)$
$[lnot (lnot p lor q) land lnot (q rightarrow r)] lor (p rightarrow r)$
$[(pland lnot q) land (q land lnot r)] lor (p rightarrow r)$
I don't know where to go from there.
logic propositional-calculus
I have been attempting to use identities to get to the answer but I am unable to get anywhere.
Here is the equation I am trying to prove tautological without using truth tables:
$[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$
And here is my attempt at using identities to work my way up. (a link to a page with a table of logical equivalence identities: http://en.wikipedia.org/wiki/Logical_equivalence)
$lnot [(p rightarrow q) land (q rightarrow r)] lor (p rightarrow r)$
$lnot [(lnot p lor q) land (q rightarrow r)] lor (p rightarrow r)$
$[lnot (lnot p lor q) land lnot (q rightarrow r)] lor (p rightarrow r)$
$[(pland lnot q) land (q land lnot r)] lor (p rightarrow r)$
I don't know where to go from there.
logic propositional-calculus
logic propositional-calculus
edited Jan 24 '15 at 21:29
Git Gud
28.7k1049100
28.7k1049100
asked Jan 24 '15 at 21:26
user11892
9828
9828
In fourth step you did not apply the nagasion properly due to nagasion sign the AND operator will change into OR operator and then the whole question will solve very easily.
– user620344
Nov 27 '18 at 18:55
add a comment |
In fourth step you did not apply the nagasion properly due to nagasion sign the AND operator will change into OR operator and then the whole question will solve very easily.
– user620344
Nov 27 '18 at 18:55
In fourth step you did not apply the nagasion properly due to nagasion sign the AND operator will change into OR operator and then the whole question will solve very easily.
– user620344
Nov 27 '18 at 18:55
In fourth step you did not apply the nagasion properly due to nagasion sign the AND operator will change into OR operator and then the whole question will solve very easily.
– user620344
Nov 27 '18 at 18:55
add a comment |
1 Answer
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In your approach, you made a mistake in the derivation on the third line, it should be:
$$[lnot (lnot p lor q) lor lnot (q rightarrow r)] lor (p rightarrow r)$$
so that your fourth line becomes:
$$[(pland lnot q) lor (q land lnot r)] lor (p rightarrow r)$$
At this point, there isn't really much to be done except for a tedious case-by-case argument ("Suppose $p$ is true. [...] Now suppose $p$ is false [...].").
Let us take a step back. When proving a tautology (which is not easily seen to be so by virtue of known theorems), it is often a more productive technique to try and prove that its negation is a contradiction. The idea behind this is that if the original $phi$ is to be a tautology, then by investigating its negation $neg phi$, we will obtain stricter and stricter conditions on $p, q, r$ to allow $neg phi$ to be true, until we get to the conclusion that there are no possibilities left, i.e. that $phi$ is a tautology.
Essentially, one converts a universal check ("All interpretations yield true") to a search for an example ("Some interpretation does not yield true") -- keeping in mind that we want the latter search to be fruitless. Basically, it's easier to find the black sheep in a flock than to count the white ones. For a more detailed account of this approach, you can search for terms like semantic tableaux or analytic tableaux.
Returning to our concrete example, let us see how this approach works out. We use repeatedly the equivalences $p to q iff neg p lor q$ and $neg (p to q) iff p land neg q$:
begin{align*}
& negbig([(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )big)\
iff& (pto q)land (qto r) land neg(p to r) \
iff& (neg p lor q) land (neg q lor r) land p land neg r
end{align*}
From the last two clauses, we see that $p$ must be true and $r$ must be false. But then for $neg p lor q$ to be true, $q$ must be true as well. Similarly, $neg q$ must be true in order for $neg q lor r$ to be true as well. Hence we need $q$ to be both true and false at the same time, a contradiction.
We conclude that $[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$ is a tautology, as desired.
add a comment |
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1 Answer
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1 Answer
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In your approach, you made a mistake in the derivation on the third line, it should be:
$$[lnot (lnot p lor q) lor lnot (q rightarrow r)] lor (p rightarrow r)$$
so that your fourth line becomes:
$$[(pland lnot q) lor (q land lnot r)] lor (p rightarrow r)$$
At this point, there isn't really much to be done except for a tedious case-by-case argument ("Suppose $p$ is true. [...] Now suppose $p$ is false [...].").
Let us take a step back. When proving a tautology (which is not easily seen to be so by virtue of known theorems), it is often a more productive technique to try and prove that its negation is a contradiction. The idea behind this is that if the original $phi$ is to be a tautology, then by investigating its negation $neg phi$, we will obtain stricter and stricter conditions on $p, q, r$ to allow $neg phi$ to be true, until we get to the conclusion that there are no possibilities left, i.e. that $phi$ is a tautology.
Essentially, one converts a universal check ("All interpretations yield true") to a search for an example ("Some interpretation does not yield true") -- keeping in mind that we want the latter search to be fruitless. Basically, it's easier to find the black sheep in a flock than to count the white ones. For a more detailed account of this approach, you can search for terms like semantic tableaux or analytic tableaux.
Returning to our concrete example, let us see how this approach works out. We use repeatedly the equivalences $p to q iff neg p lor q$ and $neg (p to q) iff p land neg q$:
begin{align*}
& negbig([(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )big)\
iff& (pto q)land (qto r) land neg(p to r) \
iff& (neg p lor q) land (neg q lor r) land p land neg r
end{align*}
From the last two clauses, we see that $p$ must be true and $r$ must be false. But then for $neg p lor q$ to be true, $q$ must be true as well. Similarly, $neg q$ must be true in order for $neg q lor r$ to be true as well. Hence we need $q$ to be both true and false at the same time, a contradiction.
We conclude that $[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$ is a tautology, as desired.
add a comment |
In your approach, you made a mistake in the derivation on the third line, it should be:
$$[lnot (lnot p lor q) lor lnot (q rightarrow r)] lor (p rightarrow r)$$
so that your fourth line becomes:
$$[(pland lnot q) lor (q land lnot r)] lor (p rightarrow r)$$
At this point, there isn't really much to be done except for a tedious case-by-case argument ("Suppose $p$ is true. [...] Now suppose $p$ is false [...].").
Let us take a step back. When proving a tautology (which is not easily seen to be so by virtue of known theorems), it is often a more productive technique to try and prove that its negation is a contradiction. The idea behind this is that if the original $phi$ is to be a tautology, then by investigating its negation $neg phi$, we will obtain stricter and stricter conditions on $p, q, r$ to allow $neg phi$ to be true, until we get to the conclusion that there are no possibilities left, i.e. that $phi$ is a tautology.
Essentially, one converts a universal check ("All interpretations yield true") to a search for an example ("Some interpretation does not yield true") -- keeping in mind that we want the latter search to be fruitless. Basically, it's easier to find the black sheep in a flock than to count the white ones. For a more detailed account of this approach, you can search for terms like semantic tableaux or analytic tableaux.
Returning to our concrete example, let us see how this approach works out. We use repeatedly the equivalences $p to q iff neg p lor q$ and $neg (p to q) iff p land neg q$:
begin{align*}
& negbig([(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )big)\
iff& (pto q)land (qto r) land neg(p to r) \
iff& (neg p lor q) land (neg q lor r) land p land neg r
end{align*}
From the last two clauses, we see that $p$ must be true and $r$ must be false. But then for $neg p lor q$ to be true, $q$ must be true as well. Similarly, $neg q$ must be true in order for $neg q lor r$ to be true as well. Hence we need $q$ to be both true and false at the same time, a contradiction.
We conclude that $[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$ is a tautology, as desired.
add a comment |
In your approach, you made a mistake in the derivation on the third line, it should be:
$$[lnot (lnot p lor q) lor lnot (q rightarrow r)] lor (p rightarrow r)$$
so that your fourth line becomes:
$$[(pland lnot q) lor (q land lnot r)] lor (p rightarrow r)$$
At this point, there isn't really much to be done except for a tedious case-by-case argument ("Suppose $p$ is true. [...] Now suppose $p$ is false [...].").
Let us take a step back. When proving a tautology (which is not easily seen to be so by virtue of known theorems), it is often a more productive technique to try and prove that its negation is a contradiction. The idea behind this is that if the original $phi$ is to be a tautology, then by investigating its negation $neg phi$, we will obtain stricter and stricter conditions on $p, q, r$ to allow $neg phi$ to be true, until we get to the conclusion that there are no possibilities left, i.e. that $phi$ is a tautology.
Essentially, one converts a universal check ("All interpretations yield true") to a search for an example ("Some interpretation does not yield true") -- keeping in mind that we want the latter search to be fruitless. Basically, it's easier to find the black sheep in a flock than to count the white ones. For a more detailed account of this approach, you can search for terms like semantic tableaux or analytic tableaux.
Returning to our concrete example, let us see how this approach works out. We use repeatedly the equivalences $p to q iff neg p lor q$ and $neg (p to q) iff p land neg q$:
begin{align*}
& negbig([(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )big)\
iff& (pto q)land (qto r) land neg(p to r) \
iff& (neg p lor q) land (neg q lor r) land p land neg r
end{align*}
From the last two clauses, we see that $p$ must be true and $r$ must be false. But then for $neg p lor q$ to be true, $q$ must be true as well. Similarly, $neg q$ must be true in order for $neg q lor r$ to be true as well. Hence we need $q$ to be both true and false at the same time, a contradiction.
We conclude that $[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$ is a tautology, as desired.
In your approach, you made a mistake in the derivation on the third line, it should be:
$$[lnot (lnot p lor q) lor lnot (q rightarrow r)] lor (p rightarrow r)$$
so that your fourth line becomes:
$$[(pland lnot q) lor (q land lnot r)] lor (p rightarrow r)$$
At this point, there isn't really much to be done except for a tedious case-by-case argument ("Suppose $p$ is true. [...] Now suppose $p$ is false [...].").
Let us take a step back. When proving a tautology (which is not easily seen to be so by virtue of known theorems), it is often a more productive technique to try and prove that its negation is a contradiction. The idea behind this is that if the original $phi$ is to be a tautology, then by investigating its negation $neg phi$, we will obtain stricter and stricter conditions on $p, q, r$ to allow $neg phi$ to be true, until we get to the conclusion that there are no possibilities left, i.e. that $phi$ is a tautology.
Essentially, one converts a universal check ("All interpretations yield true") to a search for an example ("Some interpretation does not yield true") -- keeping in mind that we want the latter search to be fruitless. Basically, it's easier to find the black sheep in a flock than to count the white ones. For a more detailed account of this approach, you can search for terms like semantic tableaux or analytic tableaux.
Returning to our concrete example, let us see how this approach works out. We use repeatedly the equivalences $p to q iff neg p lor q$ and $neg (p to q) iff p land neg q$:
begin{align*}
& negbig([(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )big)\
iff& (pto q)land (qto r) land neg(p to r) \
iff& (neg p lor q) land (neg q lor r) land p land neg r
end{align*}
From the last two clauses, we see that $p$ must be true and $r$ must be false. But then for $neg p lor q$ to be true, $q$ must be true as well. Similarly, $neg q$ must be true in order for $neg q lor r$ to be true as well. Hence we need $q$ to be both true and false at the same time, a contradiction.
We conclude that $[(prightarrow q) land (q rightarrow r)] rightarrow (p rightarrow r )$ is a tautology, as desired.
answered Jan 24 '15 at 22:56
Lord_Farin
15.5k636108
15.5k636108
add a comment |
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In fourth step you did not apply the nagasion properly due to nagasion sign the AND operator will change into OR operator and then the whole question will solve very easily.
– user620344
Nov 27 '18 at 18:55