Proving Logical equivalence predicate formulas
Clash Royale CLAN TAG#URR8PPP
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Proving the predicate formula below...
$forall x((T(x)) Rightarrow S equiv forall x(T(x) Rightarrow S) $
This is my logic, may i get some suggestions on improvement or errors if i am doing anything wrong? I just started out on proving so would appreciate it if i could get any help with it.
First scenario: Suppose S is true : Suppose $forall x ((T(x)) Rightarrow S $ to be true, all of values in T(x) must be true, therefore, $forall x(T(x) Rightarrow S) $ is true as well since it only needs one value of x to be true.
Second scenario: Suppose S is true : Suppose $forall x ((T(x)) Rightarrow S $ to be false, it means only some cases of T(x) is true. However, if this logic applies to the $forall x((Tx) Rightarrow S) $, as long as some cases come true, this predicate clause would be true, which is contradictory to the first clause.
Therefore, the two formulas are not logically equivalent.
proof-verification logic predicate-logic
edited Aug 27 at 23:27
Graham Kemp
81.1k43275
asked Aug 27 at 12:03
Alviss Min
132
add a comment |Â
up vote
1
down vote
favorite
Proving the predicate formula below...
$forall x((T(x)) Rightarrow S equiv forall x(T(x) Rightarrow S) $
This is my logic, may i get some suggestions on improvement or errors if i am doing anything wrong? I just started out on proving so would appreciate it if i could get any help with it.
First scenario: Suppose S is true : Suppose $forall x ((T(x)) Rightarrow S $ to be true, all of values in T(x) must be true, therefore, $forall x(T(x) Rightarrow S) $ is true as well since it only needs one value of x to be true.
Second scenario: Suppose S is true : Suppose $forall x ((T(x)) Rightarrow S $ to be false, it means only some cases of T(x) is true. However, if this logic applies to the $forall x((Tx) Rightarrow S) $, as long as some cases come true, this predicate clause would be true, which is contradictory to the first clause.
Therefore, the two formulas are not logically equivalent.
proof-verification logic predicate-logic
edited Aug 27 at 23:27
Graham Kemp
81.1k43275
asked Aug 27 at 12:03
Alviss Min
132
2
In fact, it turns out that $forall x (T(x) rightarrow S)$ is equivalent to $(exists x , T(x)) rightarrow S$.
– Daniel Schepler
Aug 27 at 23:27
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Proving the predicate formula below...
$forall x((T(x)) Rightarrow S equiv forall x(T(x) Rightarrow S) $
This is my logic, may i get some suggestions on improvement or errors if i am doing anything wrong? I just started out on proving so would appreciate it if i could get any help with it.
First scenario: Suppose S is true : Suppose $forall x ((T(x)) Rightarrow S $ to be true, all of values in T(x) must be true, therefore, $forall x(T(x) Rightarrow S) $ is true as well since it only needs one value of x to be true.
Second scenario: Suppose S is true : Suppose $forall x ((T(x)) Rightarrow S $ to be false, it means only some cases of T(x) is true. However, if this logic applies to the $forall x((Tx) Rightarrow S) $, as long as some cases come true, this predicate clause would be true, which is contradictory to the first clause.
Therefore, the two formulas are not logically equivalent.
proof-verification logic predicate-logic
edited Aug 27 at 23:27
Graham Kemp
81.1k43275
asked Aug 27 at 12:03
Alviss Min
132
Proving the predicate formula below...
$forall x((T(x)) Rightarrow S equiv forall x(T(x) Rightarrow S) $
This is my logic, may i get some suggestions on improvement or errors if i am doing anything wrong? I just started out on proving so would appreciate it if i could get any help with it.
First scenario: Suppose S is true : Suppose $forall x ((T(x)) Rightarrow S $ to be true, all of values in T(x) must be true, therefore, $forall x(T(x) Rightarrow S) $ is true as well since it only needs one value of x to be true.
Second scenario: Suppose S is true : Suppose $forall x ((T(x)) Rightarrow S $ to be false, it means only some cases of T(x) is true. However, if this logic applies to the $forall x((Tx) Rightarrow S) $, as long as some cases come true, this predicate clause would be true, which is contradictory to the first clause.
Therefore, the two formulas are not logically equivalent.
proof-verification logic predicate-logic
edited Aug 27 at 23:27
Graham Kemp
81.1k43275
asked Aug 27 at 12:03
Alviss Min
132
edited Aug 27 at 23:27
Graham Kemp
81.1k43275
edited Aug 27 at 23:27
Graham Kemp
81.1k43275
edited Aug 27 at 23:27
Graham Kemp
81.1k43275
81.1k43275
asked Aug 27 at 12:03
Alviss Min
132
asked Aug 27 at 12:03
Alviss Min
132
asked Aug 27 at 12:03
Alviss Min
132
132
2
In fact, it turns out that $forall x (T(x) rightarrow S)$ is equivalent to $(exists x , T(x)) rightarrow S$.
– Daniel Schepler
Aug 27 at 23:27
add a comment |Â
2
In fact, it turns out that $forall x (T(x) rightarrow S)$ is equivalent to $(exists x , T(x)) rightarrow S$.
– Daniel Schepler
Aug 27 at 23:27
2
2
In fact, it turns out that $forall x (T(x) rightarrow S)$ is equivalent to $(exists x , T(x)) rightarrow S$.
– Daniel Schepler
Aug 27 at 23:27
In fact, it turns out that $forall x (T(x) rightarrow S)$ is equivalent to $(exists x , T(x)) rightarrow S$.
– Daniel Schepler
Aug 27 at 23:27
add a comment |Â
1 Answer
1
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oldest
votes
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0
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Correct. Â The statements are not semantically equivalent, as your reasoning shows. Â
Both statements will hold in any model where $S$ is true.
In models where $S$ is false, $(forall x~T(x)) to S$ is valid exactly when not every entity makes $T$ true. Â IE: $negforall x~T(x)$
In models where $S$ is false, $forall x~(T(x)to S)$ is valid exactly when no entity makes $T$ true. Â IE: $forall x~neg T(x)$
Therefore $forall x~(T(x)to S)$ entails $(forall x~T(x)) to S$, but it is not entailed by that.
answered Aug 27 at 23:22
Graham Kemp
81.1k43275
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Correct. Â The statements are not semantically equivalent, as your reasoning shows. Â
Both statements will hold in any model where $S$ is true.
In models where $S$ is false, $(forall x~T(x)) to S$ is valid exactly when not every entity makes $T$ true. Â IE: $negforall x~T(x)$
In models where $S$ is false, $forall x~(T(x)to S)$ is valid exactly when no entity makes $T$ true. Â IE: $forall x~neg T(x)$
Therefore $forall x~(T(x)to S)$ entails $(forall x~T(x)) to S$, but it is not entailed by that.
answered Aug 27 at 23:22
Graham Kemp
81.1k43275
add a comment |Â
up vote
0
down vote
accepted
Correct. Â The statements are not semantically equivalent, as your reasoning shows. Â
Both statements will hold in any model where $S$ is true.
In models where $S$ is false, $(forall x~T(x)) to S$ is valid exactly when not every entity makes $T$ true. Â IE: $negforall x~T(x)$
In models where $S$ is false, $forall x~(T(x)to S)$ is valid exactly when no entity makes $T$ true. Â IE: $forall x~neg T(x)$
Therefore $forall x~(T(x)to S)$ entails $(forall x~T(x)) to S$, but it is not entailed by that.
answered Aug 27 at 23:22
Graham Kemp
81.1k43275
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Correct. Â The statements are not semantically equivalent, as your reasoning shows. Â
Both statements will hold in any model where $S$ is true.
In models where $S$ is false, $(forall x~T(x)) to S$ is valid exactly when not every entity makes $T$ true. Â IE: $negforall x~T(x)$
In models where $S$ is false, $forall x~(T(x)to S)$ is valid exactly when no entity makes $T$ true. Â IE: $forall x~neg T(x)$
Therefore $forall x~(T(x)to S)$ entails $(forall x~T(x)) to S$, but it is not entailed by that.
answered Aug 27 at 23:22
Graham Kemp
81.1k43275
Correct. Â The statements are not semantically equivalent, as your reasoning shows. Â
Both statements will hold in any model where $S$ is true.
In models where $S$ is false, $(forall x~T(x)) to S$ is valid exactly when not every entity makes $T$ true. Â IE: $negforall x~T(x)$
In models where $S$ is false, $forall x~(T(x)to S)$ is valid exactly when no entity makes $T$ true. Â IE: $forall x~neg T(x)$
Therefore $forall x~(T(x)to S)$ entails $(forall x~T(x)) to S$, but it is not entailed by that.
answered Aug 27 at 23:22
Graham Kemp
81.1k43275
edited Aug 27 at 23:28
answered Aug 27 at 23:22
Graham Kemp
81.1k43275
answered Aug 27 at 23:22
Graham Kemp
81.1k43275
answered Aug 27 at 23:22
Graham Kemp
81.1k43275
81.1k43275
add a comment |Â
add a comment |Â
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2
In fact, it turns out that $forall x (T(x) rightarrow S)$ is equivalent to $(exists x , T(x)) rightarrow S$.
– Daniel Schepler
Aug 27 at 23:27