Predicate logic - logically imply
Consider the following predicate formulas.
$F1: forall x exists y ( P(x) to Q(y) ).$
$F2: exists x forall y ( P(x) to Q(y) ).$
$F3: forall x P(x) to exists y Q(y).$
$F4: exists x P(x) to forall y Q(y).$
Answer the following questions with brief justification.
(a) Does $F1$ logically imply $F2$?
(b) Does $F1$ logically imply $F3$?
(c) Does $F1$ logically imply $F4$?
(d) Does $F2$ logically imply $F1$?
(e) Does $F2$ logically imply $F3$?
(f) Does $F2$ logically imply $F4$?
(g) Does $F3$ logically imply $F1$?
(h) Does $F3$ logically imply $F2$?
(i) Does $F3$ logically imply $F4$?
(j) Does $F4$ logically imply $F1$?
(k) Does $F4$ logically imply $F2$?
(l) Does $F4$ logically imply $F3$?
Logically implies really confuses me. Attempt:
Writing it in English first:
F1: We have that $x$ never satisfies $P$ or there is a $y$ that satisfies $Q$
F2: For some $x$ $P$ isn't satisfied, or $y$ always satisfies $Q$
F3: Some $x$ doesn't satisfy $P$ or some $y$ satisfies $Q$
F4: $x$ never satisfies $P$ or $y$ always satisfies Q
By definition of logically implies: A formula $F$ logically implies a formula $F'$ iff every interpretation that satisfies $F$ satisfies $F'$
So:
(a) So for $F1$ $x$ never satisfies $P$ meaning it'll always be true whereas some $x$ can satisfy P in $F2$, thus this doesn't logically imply.
(b) ... yeah I don't know how to explain any of this - any assistance even tips are appreciated I'm stumped
Apparently $b$ is true I don't get it though.
according to $F1$, $x$ never satisfies $P$ and by prepositional logic of $p to q$ is not p or q.
then $F1$ is always true whereas $F3$ there can exist a P that satisfies Q and some y that doesn't satisfy Q so every interpretation doesn't satisfy? I'm understanding this poorly.
predicate-logic
asked Dec 6 at 17:31
Tree Garen
36219
add a comment |
Consider the following predicate formulas.
$F1: forall x exists y ( P(x) to Q(y) ).$
$F2: exists x forall y ( P(x) to Q(y) ).$
$F3: forall x P(x) to exists y Q(y).$
$F4: exists x P(x) to forall y Q(y).$
Answer the following questions with brief justification.
(a) Does $F1$ logically imply $F2$?
(b) Does $F1$ logically imply $F3$?
(c) Does $F1$ logically imply $F4$?
(d) Does $F2$ logically imply $F1$?
(e) Does $F2$ logically imply $F3$?
(f) Does $F2$ logically imply $F4$?
(g) Does $F3$ logically imply $F1$?
(h) Does $F3$ logically imply $F2$?
(i) Does $F3$ logically imply $F4$?
(j) Does $F4$ logically imply $F1$?
(k) Does $F4$ logically imply $F2$?
(l) Does $F4$ logically imply $F3$?
Logically implies really confuses me. Attempt:
Writing it in English first:
F1: We have that $x$ never satisfies $P$ or there is a $y$ that satisfies $Q$
F2: For some $x$ $P$ isn't satisfied, or $y$ always satisfies $Q$
F3: Some $x$ doesn't satisfy $P$ or some $y$ satisfies $Q$
F4: $x$ never satisfies $P$ or $y$ always satisfies Q
By definition of logically implies: A formula $F$ logically implies a formula $F'$ iff every interpretation that satisfies $F$ satisfies $F'$
So:
(a) So for $F1$ $x$ never satisfies $P$ meaning it'll always be true whereas some $x$ can satisfy P in $F2$, thus this doesn't logically imply.
(b) ... yeah I don't know how to explain any of this - any assistance even tips are appreciated I'm stumped
Apparently $b$ is true I don't get it though.
according to $F1$, $x$ never satisfies $P$ and by prepositional logic of $p to q$ is not p or q.
then $F1$ is always true whereas $F3$ there can exist a P that satisfies Q and some y that doesn't satisfy Q so every interpretation doesn't satisfy? I'm understanding this poorly.
predicate-logic
asked Dec 6 at 17:31
Tree Garen
36219
add a comment |
Consider the following predicate formulas.
$F1: forall x exists y ( P(x) to Q(y) ).$
$F2: exists x forall y ( P(x) to Q(y) ).$
$F3: forall x P(x) to exists y Q(y).$
$F4: exists x P(x) to forall y Q(y).$
Answer the following questions with brief justification.
(a) Does $F1$ logically imply $F2$?
(b) Does $F1$ logically imply $F3$?
(c) Does $F1$ logically imply $F4$?
(d) Does $F2$ logically imply $F1$?
(e) Does $F2$ logically imply $F3$?
(f) Does $F2$ logically imply $F4$?
(g) Does $F3$ logically imply $F1$?
(h) Does $F3$ logically imply $F2$?
(i) Does $F3$ logically imply $F4$?
(j) Does $F4$ logically imply $F1$?
(k) Does $F4$ logically imply $F2$?
(l) Does $F4$ logically imply $F3$?
Logically implies really confuses me. Attempt:
Writing it in English first:
F1: We have that $x$ never satisfies $P$ or there is a $y$ that satisfies $Q$
F2: For some $x$ $P$ isn't satisfied, or $y$ always satisfies $Q$
F3: Some $x$ doesn't satisfy $P$ or some $y$ satisfies $Q$
F4: $x$ never satisfies $P$ or $y$ always satisfies Q
By definition of logically implies: A formula $F$ logically implies a formula $F'$ iff every interpretation that satisfies $F$ satisfies $F'$
So:
(a) So for $F1$ $x$ never satisfies $P$ meaning it'll always be true whereas some $x$ can satisfy P in $F2$, thus this doesn't logically imply.
(b) ... yeah I don't know how to explain any of this - any assistance even tips are appreciated I'm stumped
Apparently $b$ is true I don't get it though.
according to $F1$, $x$ never satisfies $P$ and by prepositional logic of $p to q$ is not p or q.
then $F1$ is always true whereas $F3$ there can exist a P that satisfies Q and some y that doesn't satisfy Q so every interpretation doesn't satisfy? I'm understanding this poorly.
predicate-logic
asked Dec 6 at 17:31
Tree Garen
36219
Consider the following predicate formulas.
$F1: forall x exists y ( P(x) to Q(y) ).$
$F2: exists x forall y ( P(x) to Q(y) ).$
$F3: forall x P(x) to exists y Q(y).$
$F4: exists x P(x) to forall y Q(y).$
Answer the following questions with brief justification.
(a) Does $F1$ logically imply $F2$?
(b) Does $F1$ logically imply $F3$?
(c) Does $F1$ logically imply $F4$?
(d) Does $F2$ logically imply $F1$?
(e) Does $F2$ logically imply $F3$?
(f) Does $F2$ logically imply $F4$?
(g) Does $F3$ logically imply $F1$?
(h) Does $F3$ logically imply $F2$?
(i) Does $F3$ logically imply $F4$?
(j) Does $F4$ logically imply $F1$?
(k) Does $F4$ logically imply $F2$?
(l) Does $F4$ logically imply $F3$?
Logically implies really confuses me. Attempt:
Writing it in English first:
F1: We have that $x$ never satisfies $P$ or there is a $y$ that satisfies $Q$
F2: For some $x$ $P$ isn't satisfied, or $y$ always satisfies $Q$
F3: Some $x$ doesn't satisfy $P$ or some $y$ satisfies $Q$
F4: $x$ never satisfies $P$ or $y$ always satisfies Q
By definition of logically implies: A formula $F$ logically implies a formula $F'$ iff every interpretation that satisfies $F$ satisfies $F'$
So:
(a) So for $F1$ $x$ never satisfies $P$ meaning it'll always be true whereas some $x$ can satisfy P in $F2$, thus this doesn't logically imply.
(b) ... yeah I don't know how to explain any of this - any assistance even tips are appreciated I'm stumped
Apparently $b$ is true I don't get it though.
according to $F1$, $x$ never satisfies $P$ and by prepositional logic of $p to q$ is not p or q.
then $F1$ is always true whereas $F3$ there can exist a P that satisfies Q and some y that doesn't satisfy Q so every interpretation doesn't satisfy? I'm understanding this poorly.
predicate-logic
predicate-logic
asked Dec 6 at 17:31
Tree Garen
36219
asked Dec 6 at 17:31
Tree Garen
36219
asked Dec 6 at 17:31
Tree Garen
36219
asked Dec 6 at 17:31
Tree Garen
36219
asked Dec 6 at 17:31
Tree Garen
36219
36219
add a comment |
add a comment |
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