$Gamma_1 cup Gamma_2$ is not satisfiable iff there exists $alpha in WFF$ such that $Gamma_1 vdash alpha$ and...
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Let $Gamma_1,Gamma_2 subseteq WFF.;$ Prove: $Gamma_1 cup Gamma_2$ is not satisfiable if and only if there exists $alpha in WFF$ such that $Gamma_1 vdash_{HPC} alpha$ and $Gamma_2 vdash_{HPC} lnot alpha$
I succeeded to prove that if there is an $alpha in WFF$ such that $Gamma_1 vdash alpha$ and $Gamma_2 vdash lnot alpha$, then $Gamma_1 cup Gamma_2$ is not satisfiable. I'm having some trouble with the other direction. If either $Gamma_1$ or $Gamma_2$ is not satisfiable, then it's pretty easy. But what if both $Gamma_1$ and $Gamma_2$ are satisfiable?
logic propositional-calculus satisfiability
asked Dec 13 '18 at 20:48
user401516user401516
92039
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Let $Gamma_1,Gamma_2 subseteq WFF.;$ Prove: $Gamma_1 cup Gamma_2$ is not satisfiable if and only if there exists $alpha in WFF$ such that $Gamma_1 vdash_{HPC} alpha$ and $Gamma_2 vdash_{HPC} lnot alpha$
I succeeded to prove that if there is an $alpha in WFF$ such that $Gamma_1 vdash alpha$ and $Gamma_2 vdash lnot alpha$, then $Gamma_1 cup Gamma_2$ is not satisfiable. I'm having some trouble with the other direction. If either $Gamma_1$ or $Gamma_2$ is not satisfiable, then it's pretty easy. But what if both $Gamma_1$ and $Gamma_2$ are satisfiable?
logic propositional-calculus satisfiability
asked Dec 13 '18 at 20:48
user401516user401516
92039
$endgroup$
add a comment |
$begingroup$
Let $Gamma_1,Gamma_2 subseteq WFF.;$ Prove: $Gamma_1 cup Gamma_2$ is not satisfiable if and only if there exists $alpha in WFF$ such that $Gamma_1 vdash_{HPC} alpha$ and $Gamma_2 vdash_{HPC} lnot alpha$
I succeeded to prove that if there is an $alpha in WFF$ such that $Gamma_1 vdash alpha$ and $Gamma_2 vdash lnot alpha$, then $Gamma_1 cup Gamma_2$ is not satisfiable. I'm having some trouble with the other direction. If either $Gamma_1$ or $Gamma_2$ is not satisfiable, then it's pretty easy. But what if both $Gamma_1$ and $Gamma_2$ are satisfiable?
logic propositional-calculus satisfiability
asked Dec 13 '18 at 20:48
user401516user401516
92039
$endgroup$
Let $Gamma_1,Gamma_2 subseteq WFF.;$ Prove: $Gamma_1 cup Gamma_2$ is not satisfiable if and only if there exists $alpha in WFF$ such that $Gamma_1 vdash_{HPC} alpha$ and $Gamma_2 vdash_{HPC} lnot alpha$
I succeeded to prove that if there is an $alpha in WFF$ such that $Gamma_1 vdash alpha$ and $Gamma_2 vdash lnot alpha$, then $Gamma_1 cup Gamma_2$ is not satisfiable. I'm having some trouble with the other direction. If either $Gamma_1$ or $Gamma_2$ is not satisfiable, then it's pretty easy. But what if both $Gamma_1$ and $Gamma_2$ are satisfiable?
logic propositional-calculus satisfiability
logic propositional-calculus satisfiability
asked Dec 13 '18 at 20:48
user401516user401516
92039
asked Dec 13 '18 at 20:48
user401516user401516
92039
asked Dec 13 '18 at 20:48
user401516user401516
92039
asked Dec 13 '18 at 20:48
user401516user401516
92039
asked Dec 13 '18 at 20:48
user401516user401516
92039
92039
add a comment |
add a comment |
1 Answer
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If $Gamma_1 cup Gamma_2$ is not satisfiable, but $Gamma_1$ and $Gamma_2$ both are, then you could consider the sentence that is the conjunction of all the sentences in $Gamma_1$ (if $Gamma_1$ is empty, then just consider any tautology). Let's call this statement $S_1$. Clearly $Gamma_1 vdash_{HPC} S_1$. Moreover, if it would be the case that $Gamma_2 not vdash_{HPC} neg S_1$, then that means that there is a valuation that sets all of the statements in $Gamma_2$ to True, but $neg S_1$ to False. But that means that that valuatin sets $S_1$ to True, and hence all sentences in $Gamma_1$ to True, meaning that this valuatin would set all sentences in $Gamma_1 cup Gamma_2$ to True, meaning that $Gamma_1 cup Gamma_2$ would be satisfiable, whch contradicts the assumption. So, it must be true that $Gamma_2 vdash neg S_1$. And so we have found our $alpha$, as desired.
answered Jan 13 at 3:46
Bram28Bram28
61.5k44792
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Since we're not given that $Gamma_1$ is finite, an application of the compactness theorem (to $Gamma_1cupGamma_2$) is needed before forming the conjunction $S_1$.
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– Andreas Blass
Jan 13 at 4:37
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@AndreasBlass Ah, yes, good point, thanks!
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– Bram28
Jan 13 at 15:59
add a comment |
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If $Gamma_1 cup Gamma_2$ is not satisfiable, but $Gamma_1$ and $Gamma_2$ both are, then you could consider the sentence that is the conjunction of all the sentences in $Gamma_1$ (if $Gamma_1$ is empty, then just consider any tautology). Let's call this statement $S_1$. Clearly $Gamma_1 vdash_{HPC} S_1$. Moreover, if it would be the case that $Gamma_2 not vdash_{HPC} neg S_1$, then that means that there is a valuation that sets all of the statements in $Gamma_2$ to True, but $neg S_1$ to False. But that means that that valuatin sets $S_1$ to True, and hence all sentences in $Gamma_1$ to True, meaning that this valuatin would set all sentences in $Gamma_1 cup Gamma_2$ to True, meaning that $Gamma_1 cup Gamma_2$ would be satisfiable, whch contradicts the assumption. So, it must be true that $Gamma_2 vdash neg S_1$. And so we have found our $alpha$, as desired.
answered Jan 13 at 3:46
Bram28Bram28
61.5k44792
$endgroup$
$begingroup$
Since we're not given that $Gamma_1$ is finite, an application of the compactness theorem (to $Gamma_1cupGamma_2$) is needed before forming the conjunction $S_1$.
$endgroup$
– Andreas Blass
Jan 13 at 4:37
$begingroup$
@AndreasBlass Ah, yes, good point, thanks!
$endgroup$
– Bram28
Jan 13 at 15:59
add a comment |
$begingroup$
If $Gamma_1 cup Gamma_2$ is not satisfiable, but $Gamma_1$ and $Gamma_2$ both are, then you could consider the sentence that is the conjunction of all the sentences in $Gamma_1$ (if $Gamma_1$ is empty, then just consider any tautology). Let's call this statement $S_1$. Clearly $Gamma_1 vdash_{HPC} S_1$. Moreover, if it would be the case that $Gamma_2 not vdash_{HPC} neg S_1$, then that means that there is a valuation that sets all of the statements in $Gamma_2$ to True, but $neg S_1$ to False. But that means that that valuatin sets $S_1$ to True, and hence all sentences in $Gamma_1$ to True, meaning that this valuatin would set all sentences in $Gamma_1 cup Gamma_2$ to True, meaning that $Gamma_1 cup Gamma_2$ would be satisfiable, whch contradicts the assumption. So, it must be true that $Gamma_2 vdash neg S_1$. And so we have found our $alpha$, as desired.
answered Jan 13 at 3:46
Bram28Bram28
61.5k44792
$endgroup$
$begingroup$
Since we're not given that $Gamma_1$ is finite, an application of the compactness theorem (to $Gamma_1cupGamma_2$) is needed before forming the conjunction $S_1$.
$endgroup$
– Andreas Blass
Jan 13 at 4:37
$begingroup$
@AndreasBlass Ah, yes, good point, thanks!
$endgroup$
– Bram28
Jan 13 at 15:59
add a comment |
$begingroup$
If $Gamma_1 cup Gamma_2$ is not satisfiable, but $Gamma_1$ and $Gamma_2$ both are, then you could consider the sentence that is the conjunction of all the sentences in $Gamma_1$ (if $Gamma_1$ is empty, then just consider any tautology). Let's call this statement $S_1$. Clearly $Gamma_1 vdash_{HPC} S_1$. Moreover, if it would be the case that $Gamma_2 not vdash_{HPC} neg S_1$, then that means that there is a valuation that sets all of the statements in $Gamma_2$ to True, but $neg S_1$ to False. But that means that that valuatin sets $S_1$ to True, and hence all sentences in $Gamma_1$ to True, meaning that this valuatin would set all sentences in $Gamma_1 cup Gamma_2$ to True, meaning that $Gamma_1 cup Gamma_2$ would be satisfiable, whch contradicts the assumption. So, it must be true that $Gamma_2 vdash neg S_1$. And so we have found our $alpha$, as desired.
answered Jan 13 at 3:46
Bram28Bram28
61.5k44792
$endgroup$
If $Gamma_1 cup Gamma_2$ is not satisfiable, but $Gamma_1$ and $Gamma_2$ both are, then you could consider the sentence that is the conjunction of all the sentences in $Gamma_1$ (if $Gamma_1$ is empty, then just consider any tautology). Let's call this statement $S_1$. Clearly $Gamma_1 vdash_{HPC} S_1$. Moreover, if it would be the case that $Gamma_2 not vdash_{HPC} neg S_1$, then that means that there is a valuation that sets all of the statements in $Gamma_2$ to True, but $neg S_1$ to False. But that means that that valuatin sets $S_1$ to True, and hence all sentences in $Gamma_1$ to True, meaning that this valuatin would set all sentences in $Gamma_1 cup Gamma_2$ to True, meaning that $Gamma_1 cup Gamma_2$ would be satisfiable, whch contradicts the assumption. So, it must be true that $Gamma_2 vdash neg S_1$. And so we have found our $alpha$, as desired.
answered Jan 13 at 3:46
Bram28Bram28
61.5k44792
answered Jan 13 at 3:46
Bram28Bram28
61.5k44792
answered Jan 13 at 3:46
Bram28Bram28
61.5k44792
answered Jan 13 at 3:46
Bram28Bram28
61.5k44792
61.5k44792
$begingroup$
Since we're not given that $Gamma_1$ is finite, an application of the compactness theorem (to $Gamma_1cupGamma_2$) is needed before forming the conjunction $S_1$.
$endgroup$
– Andreas Blass
Jan 13 at 4:37
$begingroup$
@AndreasBlass Ah, yes, good point, thanks!
$endgroup$
– Bram28
Jan 13 at 15:59
add a comment |
$begingroup$
Since we're not given that $Gamma_1$ is finite, an application of the compactness theorem (to $Gamma_1cupGamma_2$) is needed before forming the conjunction $S_1$.
$endgroup$
– Andreas Blass
Jan 13 at 4:37
$begingroup$
@AndreasBlass Ah, yes, good point, thanks!
$endgroup$
– Bram28
Jan 13 at 15:59
$begingroup$
Since we're not given that $Gamma_1$ is finite, an application of the compactness theorem (to $Gamma_1cupGamma_2$) is needed before forming the conjunction $S_1$.
$endgroup$
– Andreas Blass
Jan 13 at 4:37
$begingroup$
Since we're not given that $Gamma_1$ is finite, an application of the compactness theorem (to $Gamma_1cupGamma_2$) is needed before forming the conjunction $S_1$.
$endgroup$
– Andreas Blass
Jan 13 at 4:37
$begingroup$
@AndreasBlass Ah, yes, good point, thanks!
$endgroup$
– Bram28
Jan 13 at 15:59
$begingroup$
@AndreasBlass Ah, yes, good point, thanks!
$endgroup$
– Bram28
Jan 13 at 15:59
add a comment |
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