Prove falsity of argument schemas in predicate logic
Multi tool use
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I have to give a counterexample for the following argument schema:
∃x (Px ∧ Qx) ⊨ ∀x (Px ∨ Qx)
by definig its domain and the interpretation function, which is where I have some slight problems. On a technical level, I think I understood, what's the problem with the schema, namely, that the conclusion is false because of the universal quantifier.
But I don't really understand how to express this in an interpretational function, that proves that the first statement is true, where as the second is false.
examples-counterexamples predicate-logic
asked Dec 9 '18 at 23:15
K. MeyerK. Meyer
62
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add a comment |
$begingroup$
I have to give a counterexample for the following argument schema:
∃x (Px ∧ Qx) ⊨ ∀x (Px ∨ Qx)
by definig its domain and the interpretation function, which is where I have some slight problems. On a technical level, I think I understood, what's the problem with the schema, namely, that the conclusion is false because of the universal quantifier.
But I don't really understand how to express this in an interpretational function, that proves that the first statement is true, where as the second is false.
examples-counterexamples predicate-logic
asked Dec 9 '18 at 23:15
K. MeyerK. Meyer
62
$endgroup$
add a comment |
$begingroup$
I have to give a counterexample for the following argument schema:
∃x (Px ∧ Qx) ⊨ ∀x (Px ∨ Qx)
by definig its domain and the interpretation function, which is where I have some slight problems. On a technical level, I think I understood, what's the problem with the schema, namely, that the conclusion is false because of the universal quantifier.
But I don't really understand how to express this in an interpretational function, that proves that the first statement is true, where as the second is false.
examples-counterexamples predicate-logic
asked Dec 9 '18 at 23:15
K. MeyerK. Meyer
62
$endgroup$
I have to give a counterexample for the following argument schema:
∃x (Px ∧ Qx) ⊨ ∀x (Px ∨ Qx)
by definig its domain and the interpretation function, which is where I have some slight problems. On a technical level, I think I understood, what's the problem with the schema, namely, that the conclusion is false because of the universal quantifier.
But I don't really understand how to express this in an interpretational function, that proves that the first statement is true, where as the second is false.
examples-counterexamples predicate-logic
examples-counterexamples predicate-logic
asked Dec 9 '18 at 23:15
K. MeyerK. Meyer
62
asked Dec 9 '18 at 23:15
K. MeyerK. Meyer
62
asked Dec 9 '18 at 23:15
K. MeyerK. Meyer
62
asked Dec 9 '18 at 23:15
K. MeyerK. Meyer
62
asked Dec 9 '18 at 23:15
K. MeyerK. Meyer
62
62
add a comment |
add a comment |
1 Answer
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$begingroup$
You can set the domain to be $mathbb{R}$, so your quantifiers range over that set. And you can define the interpretation $mathcal{I}$ such that,
begin{align*}
mathcal{I}(P)(x)& triangleq x^2 = 4\
mathcal{I}(Q)(x)& triangleq x+2=4
end{align*}
Then your schema becomes,
$$exists xinmathbb{R}. (x^2=4) land (x+2=4)vDash forall xinmathbb{R}. (x^2=4) lor (x+2=4)$$
And it is trivial to show that the first is true taking $x=2$, and that the second is false choosing any $xneq 2$.
answered Dec 9 '18 at 23:54
Jorge AdrianoJorge Adriano
59146
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can set the domain to be $mathbb{R}$, so your quantifiers range over that set. And you can define the interpretation $mathcal{I}$ such that,
begin{align*}
mathcal{I}(P)(x)& triangleq x^2 = 4\
mathcal{I}(Q)(x)& triangleq x+2=4
end{align*}
Then your schema becomes,
$$exists xinmathbb{R}. (x^2=4) land (x+2=4)vDash forall xinmathbb{R}. (x^2=4) lor (x+2=4)$$
And it is trivial to show that the first is true taking $x=2$, and that the second is false choosing any $xneq 2$.
answered Dec 9 '18 at 23:54
Jorge AdrianoJorge Adriano
59146
$endgroup$
add a comment |
$begingroup$
You can set the domain to be $mathbb{R}$, so your quantifiers range over that set. And you can define the interpretation $mathcal{I}$ such that,
begin{align*}
mathcal{I}(P)(x)& triangleq x^2 = 4\
mathcal{I}(Q)(x)& triangleq x+2=4
end{align*}
Then your schema becomes,
$$exists xinmathbb{R}. (x^2=4) land (x+2=4)vDash forall xinmathbb{R}. (x^2=4) lor (x+2=4)$$
And it is trivial to show that the first is true taking $x=2$, and that the second is false choosing any $xneq 2$.
answered Dec 9 '18 at 23:54
Jorge AdrianoJorge Adriano
59146
$endgroup$
add a comment |
$begingroup$
You can set the domain to be $mathbb{R}$, so your quantifiers range over that set. And you can define the interpretation $mathcal{I}$ such that,
begin{align*}
mathcal{I}(P)(x)& triangleq x^2 = 4\
mathcal{I}(Q)(x)& triangleq x+2=4
end{align*}
Then your schema becomes,
$$exists xinmathbb{R}. (x^2=4) land (x+2=4)vDash forall xinmathbb{R}. (x^2=4) lor (x+2=4)$$
And it is trivial to show that the first is true taking $x=2$, and that the second is false choosing any $xneq 2$.
answered Dec 9 '18 at 23:54
Jorge AdrianoJorge Adriano
59146
$endgroup$
You can set the domain to be $mathbb{R}$, so your quantifiers range over that set. And you can define the interpretation $mathcal{I}$ such that,
begin{align*}
mathcal{I}(P)(x)& triangleq x^2 = 4\
mathcal{I}(Q)(x)& triangleq x+2=4
end{align*}
Then your schema becomes,
$$exists xinmathbb{R}. (x^2=4) land (x+2=4)vDash forall xinmathbb{R}. (x^2=4) lor (x+2=4)$$
And it is trivial to show that the first is true taking $x=2$, and that the second is false choosing any $xneq 2$.
answered Dec 9 '18 at 23:54
Jorge AdrianoJorge Adriano
59146
edited Dec 19 '18 at 15:09
answered Dec 9 '18 at 23:54
Jorge AdrianoJorge Adriano
59146
answered Dec 9 '18 at 23:54
Jorge AdrianoJorge Adriano
59146
answered Dec 9 '18 at 23:54
Jorge AdrianoJorge Adriano
59146
59146
add a comment |
add a comment |
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