Parentheses for quantifiers in First-Order Logic affect logical operators normally?
I'm sure this is a very simple question, but I cannot manage to find it explictly answered anywhere. Do parentheses used for quantifiers also affect logical operators normally?
For example, if parentheses affect logical operators normally, the following formula:
$forall x(P(x) land exists y(Q(y) rightarrow R(x,y)))$
Would be equivalent, in terms of its structure, to:
$P land (Q rightarrow R)$
Whereas this one:
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$
Would be equivalent to:
$P land Q rightarrow R equiv (P land Q) rightarrow R$
If it turns out that parentheses used for quantifiers do not affect the preference of "non-quantifier" logical operators, then both cases would be the same... So could someone please confirm whether I am right, and parentheses used for quantifiers affect all logical operators normally?
Thanks in advance for your time! :)
first-order-logic predicate-logic quantifiers
asked Nov 26 at 11:56
Johanovski
1
|
show 1 more comment
I'm sure this is a very simple question, but I cannot manage to find it explictly answered anywhere. Do parentheses used for quantifiers also affect logical operators normally?
For example, if parentheses affect logical operators normally, the following formula:
$forall x(P(x) land exists y(Q(y) rightarrow R(x,y)))$
Would be equivalent, in terms of its structure, to:
$P land (Q rightarrow R)$
Whereas this one:
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$
Would be equivalent to:
$P land Q rightarrow R equiv (P land Q) rightarrow R$
If it turns out that parentheses used for quantifiers do not affect the preference of "non-quantifier" logical operators, then both cases would be the same... So could someone please confirm whether I am right, and parentheses used for quantifiers affect all logical operators normally?
Thanks in advance for your time! :)
first-order-logic predicate-logic quantifiers
asked Nov 26 at 11:56
Johanovski
1
1
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$ Has unmatched parenthesis, and so is ambiguous.
– amWhy
Nov 26 at 12:02
Can you say what " parentheses affect logical operators normally," means?
– ancientmathematician
Nov 26 at 12:03
The formula shouldn't have unmatched parentheses: the outer ones embrace the whole formula from $forall x$ onwards, and, aside from the parentheses for each predicate, the $exists y$ has parentheses only surrounding the Q predicate.
– Johanovski
Nov 26 at 12:08
By that I mean whether they modify the priority of operators: meaning that $P land Q rightarrow R$ would be equivalent to $(P land Q) rightarrow R$, because $land$ has priority over $rightarrow$, but $P land (Q rightarrow R)$ would not be equivalent to the previous formula, as the explicit parentheses would modify the "normal" priority of $land$ over $rightarrow$.
– Johanovski
Nov 26 at 12:10
Yes, parenthesis used for quantifiers affect all logical operators normally.
– Bertrand Wittgenstein's Ghost
Nov 26 at 12:58
|
show 1 more comment
I'm sure this is a very simple question, but I cannot manage to find it explictly answered anywhere. Do parentheses used for quantifiers also affect logical operators normally?
For example, if parentheses affect logical operators normally, the following formula:
$forall x(P(x) land exists y(Q(y) rightarrow R(x,y)))$
Would be equivalent, in terms of its structure, to:
$P land (Q rightarrow R)$
Whereas this one:
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$
Would be equivalent to:
$P land Q rightarrow R equiv (P land Q) rightarrow R$
If it turns out that parentheses used for quantifiers do not affect the preference of "non-quantifier" logical operators, then both cases would be the same... So could someone please confirm whether I am right, and parentheses used for quantifiers affect all logical operators normally?
Thanks in advance for your time! :)
first-order-logic predicate-logic quantifiers
asked Nov 26 at 11:56
Johanovski
1
I'm sure this is a very simple question, but I cannot manage to find it explictly answered anywhere. Do parentheses used for quantifiers also affect logical operators normally?
For example, if parentheses affect logical operators normally, the following formula:
$forall x(P(x) land exists y(Q(y) rightarrow R(x,y)))$
Would be equivalent, in terms of its structure, to:
$P land (Q rightarrow R)$
Whereas this one:
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$
Would be equivalent to:
$P land Q rightarrow R equiv (P land Q) rightarrow R$
If it turns out that parentheses used for quantifiers do not affect the preference of "non-quantifier" logical operators, then both cases would be the same... So could someone please confirm whether I am right, and parentheses used for quantifiers affect all logical operators normally?
Thanks in advance for your time! :)
first-order-logic predicate-logic quantifiers
first-order-logic predicate-logic quantifiers
asked Nov 26 at 11:56
Johanovski
1
asked Nov 26 at 11:56
Johanovski
1
asked Nov 26 at 11:56
Johanovski
1
asked Nov 26 at 11:56
Johanovski
1
asked Nov 26 at 11:56
Johanovski
1
1
1
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$ Has unmatched parenthesis, and so is ambiguous.
– amWhy
Nov 26 at 12:02
Can you say what " parentheses affect logical operators normally," means?
– ancientmathematician
Nov 26 at 12:03
The formula shouldn't have unmatched parentheses: the outer ones embrace the whole formula from $forall x$ onwards, and, aside from the parentheses for each predicate, the $exists y$ has parentheses only surrounding the Q predicate.
– Johanovski
Nov 26 at 12:08
By that I mean whether they modify the priority of operators: meaning that $P land Q rightarrow R$ would be equivalent to $(P land Q) rightarrow R$, because $land$ has priority over $rightarrow$, but $P land (Q rightarrow R)$ would not be equivalent to the previous formula, as the explicit parentheses would modify the "normal" priority of $land$ over $rightarrow$.
– Johanovski
Nov 26 at 12:10
Yes, parenthesis used for quantifiers affect all logical operators normally.
– Bertrand Wittgenstein's Ghost
Nov 26 at 12:58
|
show 1 more comment
1
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$ Has unmatched parenthesis, and so is ambiguous.
– amWhy
Nov 26 at 12:02
Can you say what " parentheses affect logical operators normally," means?
– ancientmathematician
Nov 26 at 12:03
The formula shouldn't have unmatched parentheses: the outer ones embrace the whole formula from $forall x$ onwards, and, aside from the parentheses for each predicate, the $exists y$ has parentheses only surrounding the Q predicate.
– Johanovski
Nov 26 at 12:08
By that I mean whether they modify the priority of operators: meaning that $P land Q rightarrow R$ would be equivalent to $(P land Q) rightarrow R$, because $land$ has priority over $rightarrow$, but $P land (Q rightarrow R)$ would not be equivalent to the previous formula, as the explicit parentheses would modify the "normal" priority of $land$ over $rightarrow$.
– Johanovski
Nov 26 at 12:10
Yes, parenthesis used for quantifiers affect all logical operators normally.
– Bertrand Wittgenstein's Ghost
Nov 26 at 12:58
1
1
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$ Has unmatched parenthesis, and so is ambiguous.
– amWhy
Nov 26 at 12:02
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$ Has unmatched parenthesis, and so is ambiguous.
– amWhy
Nov 26 at 12:02
Can you say what " parentheses affect logical operators normally," means?
– ancientmathematician
Nov 26 at 12:03
Can you say what " parentheses affect logical operators normally," means?
– ancientmathematician
Nov 26 at 12:03
The formula shouldn't have unmatched parentheses: the outer ones embrace the whole formula from $forall x$ onwards, and, aside from the parentheses for each predicate, the $exists y$ has parentheses only surrounding the Q predicate.
– Johanovski
Nov 26 at 12:08
The formula shouldn't have unmatched parentheses: the outer ones embrace the whole formula from $forall x$ onwards, and, aside from the parentheses for each predicate, the $exists y$ has parentheses only surrounding the Q predicate.
– Johanovski
Nov 26 at 12:08
By that I mean whether they modify the priority of operators: meaning that $P land Q rightarrow R$ would be equivalent to $(P land Q) rightarrow R$, because $land$ has priority over $rightarrow$, but $P land (Q rightarrow R)$ would not be equivalent to the previous formula, as the explicit parentheses would modify the "normal" priority of $land$ over $rightarrow$.
– Johanovski
Nov 26 at 12:10
By that I mean whether they modify the priority of operators: meaning that $P land Q rightarrow R$ would be equivalent to $(P land Q) rightarrow R$, because $land$ has priority over $rightarrow$, but $P land (Q rightarrow R)$ would not be equivalent to the previous formula, as the explicit parentheses would modify the "normal" priority of $land$ over $rightarrow$.
– Johanovski
Nov 26 at 12:10
Yes, parenthesis used for quantifiers affect all logical operators normally.
– Bertrand Wittgenstein's Ghost
Nov 26 at 12:58
Yes, parenthesis used for quantifiers affect all logical operators normally.
– Bertrand Wittgenstein's Ghost
Nov 26 at 12:58
|
show 1 more comment
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1
$forall x(P(x) land exists y(Q(y)) rightarrow R(x,y))$ Has unmatched parenthesis, and so is ambiguous.
– amWhy
Nov 26 at 12:02
Can you say what " parentheses affect logical operators normally," means?
– ancientmathematician
Nov 26 at 12:03
The formula shouldn't have unmatched parentheses: the outer ones embrace the whole formula from $forall x$ onwards, and, aside from the parentheses for each predicate, the $exists y$ has parentheses only surrounding the Q predicate.
– Johanovski
Nov 26 at 12:08
By that I mean whether they modify the priority of operators: meaning that $P land Q rightarrow R$ would be equivalent to $(P land Q) rightarrow R$, because $land$ has priority over $rightarrow$, but $P land (Q rightarrow R)$ would not be equivalent to the previous formula, as the explicit parentheses would modify the "normal" priority of $land$ over $rightarrow$.
– Johanovski
Nov 26 at 12:10
Yes, parenthesis used for quantifiers affect all logical operators normally.
– Bertrand Wittgenstein's Ghost
Nov 26 at 12:58