Mixing
Is it true that $((pland q rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$?
$begingroup$
Is this a sound inference rule?
$$((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$$
So far I've rewritten it to
$$((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$$
It looks similar to the syllogism rule, but I'm not sure. Any help?
logic propositional-calculus
edited Jan 17 at 21:48
Arturo Magidin
264k34587915
asked Jan 17 at 18:09
user635758user635758
112
$endgroup$
add a comment |
$begingroup$
Is this a sound inference rule?
$$((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$$
So far I've rewritten it to
$$((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$$
It looks similar to the syllogism rule, but I'm not sure. Any help?
logic propositional-calculus
edited Jan 17 at 21:48
Arturo Magidin
264k34587915
asked Jan 17 at 18:09
user635758user635758
112
$endgroup$
3
$begingroup$
$(pto r)landneg(pto r)$ is a contradiction. The usual proof laws allow you to infer anything whatever from a contradiction.
$endgroup$
– Adrian Keister
Jan 17 at 18:15
$begingroup$
$((pland q)to r)land lnot(pto r) equiv (pto (qto r)) land lnot(pto r)$ which is not equivalent to $((p → r) ∧ ¬(p → r))$.
$endgroup$
– jordan_glen
Jan 17 at 18:28
1
$begingroup$
Can you do truth tables?
$endgroup$
– GEdgar
Jan 17 at 19:45
$begingroup$
@GEdgar, yes, do you have something specific in mind?
$endgroup$
– user635758
Jan 17 at 20:13
add a comment |
$begingroup$
Is this a sound inference rule?
$$((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$$
So far I've rewritten it to
$$((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$$
It looks similar to the syllogism rule, but I'm not sure. Any help?
logic propositional-calculus
edited Jan 17 at 21:48
Arturo Magidin
264k34587915
asked Jan 17 at 18:09
user635758user635758
112
$endgroup$
Is this a sound inference rule?
$$((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$$
So far I've rewritten it to
$$((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$$
It looks similar to the syllogism rule, but I'm not sure. Any help?
logic propositional-calculus
logic propositional-calculus
edited Jan 17 at 21:48
Arturo Magidin
264k34587915
asked Jan 17 at 18:09
user635758user635758
112
edited Jan 17 at 21:48
Arturo Magidin
264k34587915
asked Jan 17 at 18:09
user635758user635758
112
edited Jan 17 at 21:48
Arturo Magidin
264k34587915
edited Jan 17 at 21:48
Arturo Magidin
264k34587915
edited Jan 17 at 21:48
Arturo Magidin
264k34587915
264k34587915
asked Jan 17 at 18:09
user635758user635758
112
asked Jan 17 at 18:09
user635758user635758
112
asked Jan 17 at 18:09
user635758user635758
112
112
3
$begingroup$
$(pto r)landneg(pto r)$ is a contradiction. The usual proof laws allow you to infer anything whatever from a contradiction.
$endgroup$
– Adrian Keister
Jan 17 at 18:15
$begingroup$
$((pland q)to r)land lnot(pto r) equiv (pto (qto r)) land lnot(pto r)$ which is not equivalent to $((p → r) ∧ ¬(p → r))$.
$endgroup$
– jordan_glen
Jan 17 at 18:28
1
$begingroup$
Can you do truth tables?
$endgroup$
– GEdgar
Jan 17 at 19:45
$begingroup$
@GEdgar, yes, do you have something specific in mind?
$endgroup$
– user635758
Jan 17 at 20:13
add a comment |
3
$begingroup$
$(pto r)landneg(pto r)$ is a contradiction. The usual proof laws allow you to infer anything whatever from a contradiction.
$endgroup$
– Adrian Keister
Jan 17 at 18:15
$begingroup$
$((pland q)to r)land lnot(pto r) equiv (pto (qto r)) land lnot(pto r)$ which is not equivalent to $((p → r) ∧ ¬(p → r))$.
$endgroup$
– jordan_glen
Jan 17 at 18:28
1
$begingroup$
Can you do truth tables?
$endgroup$
– GEdgar
Jan 17 at 19:45
$begingroup$
@GEdgar, yes, do you have something specific in mind?
$endgroup$
– user635758
Jan 17 at 20:13
3
3
$begingroup$
$(pto r)landneg(pto r)$ is a contradiction. The usual proof laws allow you to infer anything whatever from a contradiction.
$endgroup$
– Adrian Keister
Jan 17 at 18:15
$begingroup$
$(pto r)landneg(pto r)$ is a contradiction. The usual proof laws allow you to infer anything whatever from a contradiction.
$endgroup$
– Adrian Keister
Jan 17 at 18:15
$begingroup$
$((pland q)to r)land lnot(pto r) equiv (pto (qto r)) land lnot(pto r)$ which is not equivalent to $((p → r) ∧ ¬(p → r))$.
$endgroup$
– jordan_glen
Jan 17 at 18:28
$begingroup$
$((pland q)to r)land lnot(pto r) equiv (pto (qto r)) land lnot(pto r)$ which is not equivalent to $((p → r) ∧ ¬(p → r))$.
$endgroup$
– jordan_glen
Jan 17 at 18:28
1
1
$begingroup$
Can you do truth tables?
$endgroup$
– GEdgar
Jan 17 at 19:45
$begingroup$
Can you do truth tables?
$endgroup$
– GEdgar
Jan 17 at 19:45
$begingroup$
@GEdgar, yes, do you have something specific in mind?
$endgroup$
– user635758
Jan 17 at 20:13
$begingroup$
@GEdgar, yes, do you have something specific in mind?
$endgroup$
– user635758
Jan 17 at 20:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
@GEdgar's suggestion was spot on. If you write a truth tabe to consider $2^3 = 8$ possible truth-value assignments for $(p, q, r)$, you will see that your given expression is, in fact, a tautology, meaning that no matter what truth values you assign to $p, q$ and $r$, the propostion will always be true:
Hence, indeed, the implication is a valid and sound inference rule.
answered Jan 17 at 20:12
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
Is this possible to prove using some rules of inferences?
$endgroup$
– user635758
Jan 17 at 20:16
$begingroup$
By "rules of inference" do you mean something like a natural deduction proof, or do you mean something more like using identities to rewrite the expression until you get T?
$endgroup$
– Daniel Schepler
Jan 17 at 20:32
$begingroup$
@DanielSchepler, I was thinking about if there is possible to rewrite the expression into something fitting any of these rules geeksforgeeks.org/wp-content/ql-cache/…
$endgroup$
– user635758
Jan 17 at 20:39
$begingroup$
@user635758 The "rules of inference" column there definitely looks in the spirit of natural deduction, I can write out an answer using a system I'm more familiar with if you like.
$endgroup$
– Daniel Schepler
Jan 17 at 20:43
$begingroup$
This is weird. I mean, yes, it is a tautology; and I can even verify from first principles that you cannot make the implication false. But I still balk at it; it seems to say that if two things imply a third and the first by itself does not, then the second by itself does. But I keep thinking about situations in which two statements imply a third in conjunction, but neither of them does by itself. E.g., $p=“aleq b”$, $q=“aneq b”$, $r=“alt b”$. Certainly, $aleq bland aneq brightarrow alt b$, and $neg(aleq b rightarrow alt b)$, but $neg (aneq brightarrow alt b)$.
$endgroup$
– Arturo Magidin
Jan 18 at 12:47
|
show 2 more comments
$begingroup$
Not using your rules of inference, but mine...
[ 1]TO PROVE $((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$
[ 2]$quad$ ASSUME $((p land q) rightarrow r) land lnot(p rightarrow r)$
[ 3]$quad$ TO PROVE $q rightarrow r$
[ 4]$qquad$ ASSUME $q$
[ 5]$qquad$ TO PROVE $r$
[ 6]$qquadqquad$ $lnot(p rightarrow r)qquad$ by [2]
[ 7]$qquadqquad$ $lnot(lnot p lor r)qquad$ by [6], definition of $rightarrow$
[ 8]$qquadqquad$ $p land lnot rqquad$ by [7], DeMorgan
[ 9]$qquadqquad$ $pqquad$ by [8]
[10]$qquadqquad$ $p land qqquad$ by [9],[4]
[11]$qquadqquad$ $(pland q)rightarrow rqquad$ by [2]
[12]$qquadqquad$ $rqquad$ by [11],[10], modus ponens, establishes [5]
[13]$qquad$ $qrightarrow rqquad$ by [4]...[12], establishes [3]
[14]$ $ $((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$, by [2]...[13], establishes [1]
answered Jan 17 at 21:13
GEdgarGEdgar
62.6k267171
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
@GEdgar's suggestion was spot on. If you write a truth tabe to consider $2^3 = 8$ possible truth-value assignments for $(p, q, r)$, you will see that your given expression is, in fact, a tautology, meaning that no matter what truth values you assign to $p, q$ and $r$, the propostion will always be true:
Hence, indeed, the implication is a valid and sound inference rule.
answered Jan 17 at 20:12
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
Is this possible to prove using some rules of inferences?
$endgroup$
– user635758
Jan 17 at 20:16
$begingroup$
By "rules of inference" do you mean something like a natural deduction proof, or do you mean something more like using identities to rewrite the expression until you get T?
$endgroup$
– Daniel Schepler
Jan 17 at 20:32
$begingroup$
@DanielSchepler, I was thinking about if there is possible to rewrite the expression into something fitting any of these rules geeksforgeeks.org/wp-content/ql-cache/…
$endgroup$
– user635758
Jan 17 at 20:39
$begingroup$
@user635758 The "rules of inference" column there definitely looks in the spirit of natural deduction, I can write out an answer using a system I'm more familiar with if you like.
$endgroup$
– Daniel Schepler
Jan 17 at 20:43
$begingroup$
This is weird. I mean, yes, it is a tautology; and I can even verify from first principles that you cannot make the implication false. But I still balk at it; it seems to say that if two things imply a third and the first by itself does not, then the second by itself does. But I keep thinking about situations in which two statements imply a third in conjunction, but neither of them does by itself. E.g., $p=“aleq b”$, $q=“aneq b”$, $r=“alt b”$. Certainly, $aleq bland aneq brightarrow alt b$, and $neg(aleq b rightarrow alt b)$, but $neg (aneq brightarrow alt b)$.
$endgroup$
– Arturo Magidin
Jan 18 at 12:47
|
show 2 more comments
$begingroup$
@GEdgar's suggestion was spot on. If you write a truth tabe to consider $2^3 = 8$ possible truth-value assignments for $(p, q, r)$, you will see that your given expression is, in fact, a tautology, meaning that no matter what truth values you assign to $p, q$ and $r$, the propostion will always be true:
Hence, indeed, the implication is a valid and sound inference rule.
answered Jan 17 at 20:12
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
Is this possible to prove using some rules of inferences?
$endgroup$
– user635758
Jan 17 at 20:16
$begingroup$
By "rules of inference" do you mean something like a natural deduction proof, or do you mean something more like using identities to rewrite the expression until you get T?
$endgroup$
– Daniel Schepler
Jan 17 at 20:32
$begingroup$
@DanielSchepler, I was thinking about if there is possible to rewrite the expression into something fitting any of these rules geeksforgeeks.org/wp-content/ql-cache/…
$endgroup$
– user635758
Jan 17 at 20:39
$begingroup$
@user635758 The "rules of inference" column there definitely looks in the spirit of natural deduction, I can write out an answer using a system I'm more familiar with if you like.
$endgroup$
– Daniel Schepler
Jan 17 at 20:43
$begingroup$
This is weird. I mean, yes, it is a tautology; and I can even verify from first principles that you cannot make the implication false. But I still balk at it; it seems to say that if two things imply a third and the first by itself does not, then the second by itself does. But I keep thinking about situations in which two statements imply a third in conjunction, but neither of them does by itself. E.g., $p=“aleq b”$, $q=“aneq b”$, $r=“alt b”$. Certainly, $aleq bland aneq brightarrow alt b$, and $neg(aleq b rightarrow alt b)$, but $neg (aneq brightarrow alt b)$.
$endgroup$
– Arturo Magidin
Jan 18 at 12:47
|
show 2 more comments
$begingroup$
@GEdgar's suggestion was spot on. If you write a truth tabe to consider $2^3 = 8$ possible truth-value assignments for $(p, q, r)$, you will see that your given expression is, in fact, a tautology, meaning that no matter what truth values you assign to $p, q$ and $r$, the propostion will always be true:
Hence, indeed, the implication is a valid and sound inference rule.
answered Jan 17 at 20:12
jordan_glenjordan_glen
1
$endgroup$
@GEdgar's suggestion was spot on. If you write a truth tabe to consider $2^3 = 8$ possible truth-value assignments for $(p, q, r)$, you will see that your given expression is, in fact, a tautology, meaning that no matter what truth values you assign to $p, q$ and $r$, the propostion will always be true:
Hence, indeed, the implication is a valid and sound inference rule.
answered Jan 17 at 20:12
jordan_glenjordan_glen
1
edited Jan 17 at 20:16
answered Jan 17 at 20:12
jordan_glenjordan_glen
1
answered Jan 17 at 20:12
jordan_glenjordan_glen
1
answered Jan 17 at 20:12
jordan_glenjordan_glen
1
1
$begingroup$
Is this possible to prove using some rules of inferences?
$endgroup$
– user635758
Jan 17 at 20:16
$begingroup$
By "rules of inference" do you mean something like a natural deduction proof, or do you mean something more like using identities to rewrite the expression until you get T?
$endgroup$
– Daniel Schepler
Jan 17 at 20:32
$begingroup$
@DanielSchepler, I was thinking about if there is possible to rewrite the expression into something fitting any of these rules geeksforgeeks.org/wp-content/ql-cache/…
$endgroup$
– user635758
Jan 17 at 20:39
$begingroup$
@user635758 The "rules of inference" column there definitely looks in the spirit of natural deduction, I can write out an answer using a system I'm more familiar with if you like.
$endgroup$
– Daniel Schepler
Jan 17 at 20:43
$begingroup$
This is weird. I mean, yes, it is a tautology; and I can even verify from first principles that you cannot make the implication false. But I still balk at it; it seems to say that if two things imply a third and the first by itself does not, then the second by itself does. But I keep thinking about situations in which two statements imply a third in conjunction, but neither of them does by itself. E.g., $p=“aleq b”$, $q=“aneq b”$, $r=“alt b”$. Certainly, $aleq bland aneq brightarrow alt b$, and $neg(aleq b rightarrow alt b)$, but $neg (aneq brightarrow alt b)$.
$endgroup$
– Arturo Magidin
Jan 18 at 12:47
|
show 2 more comments
$begingroup$
Is this possible to prove using some rules of inferences?
$endgroup$
– user635758
Jan 17 at 20:16
$begingroup$
By "rules of inference" do you mean something like a natural deduction proof, or do you mean something more like using identities to rewrite the expression until you get T?
$endgroup$
– Daniel Schepler
Jan 17 at 20:32
$begingroup$
@DanielSchepler, I was thinking about if there is possible to rewrite the expression into something fitting any of these rules geeksforgeeks.org/wp-content/ql-cache/…
$endgroup$
– user635758
Jan 17 at 20:39
$begingroup$
@user635758 The "rules of inference" column there definitely looks in the spirit of natural deduction, I can write out an answer using a system I'm more familiar with if you like.
$endgroup$
– Daniel Schepler
Jan 17 at 20:43
$begingroup$
This is weird. I mean, yes, it is a tautology; and I can even verify from first principles that you cannot make the implication false. But I still balk at it; it seems to say that if two things imply a third and the first by itself does not, then the second by itself does. But I keep thinking about situations in which two statements imply a third in conjunction, but neither of them does by itself. E.g., $p=“aleq b”$, $q=“aneq b”$, $r=“alt b”$. Certainly, $aleq bland aneq brightarrow alt b$, and $neg(aleq b rightarrow alt b)$, but $neg (aneq brightarrow alt b)$.
$endgroup$
– Arturo Magidin
Jan 18 at 12:47
$begingroup$
Is this possible to prove using some rules of inferences?
$endgroup$
– user635758
Jan 17 at 20:16
$begingroup$
Is this possible to prove using some rules of inferences?
$endgroup$
– user635758
Jan 17 at 20:16
$begingroup$
By "rules of inference" do you mean something like a natural deduction proof, or do you mean something more like using identities to rewrite the expression until you get T?
$endgroup$
– Daniel Schepler
Jan 17 at 20:32
$begingroup$
By "rules of inference" do you mean something like a natural deduction proof, or do you mean something more like using identities to rewrite the expression until you get T?
$endgroup$
– Daniel Schepler
Jan 17 at 20:32
$begingroup$
@DanielSchepler, I was thinking about if there is possible to rewrite the expression into something fitting any of these rules geeksforgeeks.org/wp-content/ql-cache/…
$endgroup$
– user635758
Jan 17 at 20:39
$begingroup$
@DanielSchepler, I was thinking about if there is possible to rewrite the expression into something fitting any of these rules geeksforgeeks.org/wp-content/ql-cache/…
$endgroup$
– user635758
Jan 17 at 20:39
$begingroup$
@user635758 The "rules of inference" column there definitely looks in the spirit of natural deduction, I can write out an answer using a system I'm more familiar with if you like.
$endgroup$
– Daniel Schepler
Jan 17 at 20:43
$begingroup$
@user635758 The "rules of inference" column there definitely looks in the spirit of natural deduction, I can write out an answer using a system I'm more familiar with if you like.
$endgroup$
– Daniel Schepler
Jan 17 at 20:43
$begingroup$
This is weird. I mean, yes, it is a tautology; and I can even verify from first principles that you cannot make the implication false. But I still balk at it; it seems to say that if two things imply a third and the first by itself does not, then the second by itself does. But I keep thinking about situations in which two statements imply a third in conjunction, but neither of them does by itself. E.g., $p=“aleq b”$, $q=“aneq b”$, $r=“alt b”$. Certainly, $aleq bland aneq brightarrow alt b$, and $neg(aleq b rightarrow alt b)$, but $neg (aneq brightarrow alt b)$.
$endgroup$
– Arturo Magidin
Jan 18 at 12:47
$begingroup$
This is weird. I mean, yes, it is a tautology; and I can even verify from first principles that you cannot make the implication false. But I still balk at it; it seems to say that if two things imply a third and the first by itself does not, then the second by itself does. But I keep thinking about situations in which two statements imply a third in conjunction, but neither of them does by itself. E.g., $p=“aleq b”$, $q=“aneq b”$, $r=“alt b”$. Certainly, $aleq bland aneq brightarrow alt b$, and $neg(aleq b rightarrow alt b)$, but $neg (aneq brightarrow alt b)$.
$endgroup$
– Arturo Magidin
Jan 18 at 12:47
|
show 2 more comments
$begingroup$
Not using your rules of inference, but mine...
[ 1]TO PROVE $((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$
[ 2]$quad$ ASSUME $((p land q) rightarrow r) land lnot(p rightarrow r)$
[ 3]$quad$ TO PROVE $q rightarrow r$
[ 4]$qquad$ ASSUME $q$
[ 5]$qquad$ TO PROVE $r$
[ 6]$qquadqquad$ $lnot(p rightarrow r)qquad$ by [2]
[ 7]$qquadqquad$ $lnot(lnot p lor r)qquad$ by [6], definition of $rightarrow$
[ 8]$qquadqquad$ $p land lnot rqquad$ by [7], DeMorgan
[ 9]$qquadqquad$ $pqquad$ by [8]
[10]$qquadqquad$ $p land qqquad$ by [9],[4]
[11]$qquadqquad$ $(pland q)rightarrow rqquad$ by [2]
[12]$qquadqquad$ $rqquad$ by [11],[10], modus ponens, establishes [5]
[13]$qquad$ $qrightarrow rqquad$ by [4]...[12], establishes [3]
[14]$ $ $((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$, by [2]...[13], establishes [1]
answered Jan 17 at 21:13
GEdgarGEdgar
62.6k267171
$endgroup$
add a comment |
$begingroup$
Not using your rules of inference, but mine...
[ 1]TO PROVE $((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$
[ 2]$quad$ ASSUME $((p land q) rightarrow r) land lnot(p rightarrow r)$
[ 3]$quad$ TO PROVE $q rightarrow r$
[ 4]$qquad$ ASSUME $q$
[ 5]$qquad$ TO PROVE $r$
[ 6]$qquadqquad$ $lnot(p rightarrow r)qquad$ by [2]
[ 7]$qquadqquad$ $lnot(lnot p lor r)qquad$ by [6], definition of $rightarrow$
[ 8]$qquadqquad$ $p land lnot rqquad$ by [7], DeMorgan
[ 9]$qquadqquad$ $pqquad$ by [8]
[10]$qquadqquad$ $p land qqquad$ by [9],[4]
[11]$qquadqquad$ $(pland q)rightarrow rqquad$ by [2]
[12]$qquadqquad$ $rqquad$ by [11],[10], modus ponens, establishes [5]
[13]$qquad$ $qrightarrow rqquad$ by [4]...[12], establishes [3]
[14]$ $ $((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$, by [2]...[13], establishes [1]
answered Jan 17 at 21:13
GEdgarGEdgar
62.6k267171
$endgroup$
add a comment |
$begingroup$
Not using your rules of inference, but mine...
[ 1]TO PROVE $((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$
[ 2]$quad$ ASSUME $((p land q) rightarrow r) land lnot(p rightarrow r)$
[ 3]$quad$ TO PROVE $q rightarrow r$
[ 4]$qquad$ ASSUME $q$
[ 5]$qquad$ TO PROVE $r$
[ 6]$qquadqquad$ $lnot(p rightarrow r)qquad$ by [2]
[ 7]$qquadqquad$ $lnot(lnot p lor r)qquad$ by [6], definition of $rightarrow$
[ 8]$qquadqquad$ $p land lnot rqquad$ by [7], DeMorgan
[ 9]$qquadqquad$ $pqquad$ by [8]
[10]$qquadqquad$ $p land qqquad$ by [9],[4]
[11]$qquadqquad$ $(pland q)rightarrow rqquad$ by [2]
[12]$qquadqquad$ $rqquad$ by [11],[10], modus ponens, establishes [5]
[13]$qquad$ $qrightarrow rqquad$ by [4]...[12], establishes [3]
[14]$ $ $((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$, by [2]...[13], establishes [1]
answered Jan 17 at 21:13
GEdgarGEdgar
62.6k267171
$endgroup$
Not using your rules of inference, but mine...
[ 1]TO PROVE $((p land q) rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$
[ 2]$quad$ ASSUME $((p land q) rightarrow r) land lnot(p rightarrow r)$
[ 3]$quad$ TO PROVE $q rightarrow r$
[ 4]$qquad$ ASSUME $q$
[ 5]$qquad$ TO PROVE $r$
[ 6]$qquadqquad$ $lnot(p rightarrow r)qquad$ by [2]
[ 7]$qquadqquad$ $lnot(lnot p lor r)qquad$ by [6], definition of $rightarrow$
[ 8]$qquadqquad$ $p land lnot rqquad$ by [7], DeMorgan
[ 9]$qquadqquad$ $pqquad$ by [8]
[10]$qquadqquad$ $p land qqquad$ by [9],[4]
[11]$qquadqquad$ $(pland q)rightarrow rqquad$ by [2]
[12]$qquadqquad$ $rqquad$ by [11],[10], modus ponens, establishes [5]
[13]$qquad$ $qrightarrow rqquad$ by [4]...[12], establishes [3]
[14]$ $ $((p rightarrow r) land lnot(p rightarrow r)) rightarrow (q rightarrow r)$, by [2]...[13], establishes [1]
answered Jan 17 at 21:13
GEdgarGEdgar
62.6k267171
answered Jan 17 at 21:13
GEdgarGEdgar
62.6k267171
answered Jan 17 at 21:13
GEdgarGEdgar
62.6k267171
answered Jan 17 at 21:13
GEdgarGEdgar
62.6k267171
62.6k267171
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3
$begingroup$
$(pto r)landneg(pto r)$ is a contradiction. The usual proof laws allow you to infer anything whatever from a contradiction.
$endgroup$
– Adrian Keister
Jan 17 at 18:15
$begingroup$
$((pland q)to r)land lnot(pto r) equiv (pto (qto r)) land lnot(pto r)$ which is not equivalent to $((p → r) ∧ ¬(p → r))$.
$endgroup$
– jordan_glen
Jan 17 at 18:28
1
$begingroup$
Can you do truth tables?
$endgroup$
– GEdgar
Jan 17 at 19:45
$begingroup$
@GEdgar, yes, do you have something specific in mind?
$endgroup$
– user635758
Jan 17 at 20:13