Mixing
If $t leq s$ and $t neq s$, why $t' land s neq 0$
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In boolean algebra, why if $t leq s$ and $t neq s$, why $t' land s neq 0$, where ' is the complement. I see it if I use that all boolean algebras are ${0,1}^n$ but I want to have a proof using only the axioms.
abstract-algebra boolean-algebra
asked Jan 16 at 6:31
user2316602user2316602
286
$endgroup$
add a comment |
$begingroup$
In boolean algebra, why if $t leq s$ and $t neq s$, why $t' land s neq 0$, where ' is the complement. I see it if I use that all boolean algebras are ${0,1}^n$ but I want to have a proof using only the axioms.
abstract-algebra boolean-algebra
asked Jan 16 at 6:31
user2316602user2316602
286
$endgroup$
2
$begingroup$
It is not true that all Boolean algebras are ${0,1}^n$, for some natural $n$. That is the case for finite Boolean algebras. In general, it isn't even the case that any Boolean algebra is of the type ${0,1}^X$ for some (finite or infinite) set $X$: take, for example the algebra of finite-cofinite subsets of a set. A simple reasoning based on Cantor's theorem on cardinals allows to conclude that the Boolean algebra of finite-cofinite subsets of a countably infinite set doesn't have that shape.
$endgroup$
– amrsa
Jan 16 at 17:48
add a comment |
$begingroup$
In boolean algebra, why if $t leq s$ and $t neq s$, why $t' land s neq 0$, where ' is the complement. I see it if I use that all boolean algebras are ${0,1}^n$ but I want to have a proof using only the axioms.
abstract-algebra boolean-algebra
asked Jan 16 at 6:31
user2316602user2316602
286
$endgroup$
In boolean algebra, why if $t leq s$ and $t neq s$, why $t' land s neq 0$, where ' is the complement. I see it if I use that all boolean algebras are ${0,1}^n$ but I want to have a proof using only the axioms.
abstract-algebra boolean-algebra
abstract-algebra boolean-algebra
asked Jan 16 at 6:31
user2316602user2316602
286
asked Jan 16 at 6:31
user2316602user2316602
286
edited Jan 16 at 6:35
user2316602
asked Jan 16 at 6:31
user2316602user2316602
286
asked Jan 16 at 6:31
user2316602user2316602
286
asked Jan 16 at 6:31
user2316602user2316602
286
286
2
$begingroup$
It is not true that all Boolean algebras are ${0,1}^n$, for some natural $n$. That is the case for finite Boolean algebras. In general, it isn't even the case that any Boolean algebra is of the type ${0,1}^X$ for some (finite or infinite) set $X$: take, for example the algebra of finite-cofinite subsets of a set. A simple reasoning based on Cantor's theorem on cardinals allows to conclude that the Boolean algebra of finite-cofinite subsets of a countably infinite set doesn't have that shape.
$endgroup$
– amrsa
Jan 16 at 17:48
add a comment |
2
$begingroup$
It is not true that all Boolean algebras are ${0,1}^n$, for some natural $n$. That is the case for finite Boolean algebras. In general, it isn't even the case that any Boolean algebra is of the type ${0,1}^X$ for some (finite or infinite) set $X$: take, for example the algebra of finite-cofinite subsets of a set. A simple reasoning based on Cantor's theorem on cardinals allows to conclude that the Boolean algebra of finite-cofinite subsets of a countably infinite set doesn't have that shape.
$endgroup$
– amrsa
Jan 16 at 17:48
2
2
$begingroup$
It is not true that all Boolean algebras are ${0,1}^n$, for some natural $n$. That is the case for finite Boolean algebras. In general, it isn't even the case that any Boolean algebra is of the type ${0,1}^X$ for some (finite or infinite) set $X$: take, for example the algebra of finite-cofinite subsets of a set. A simple reasoning based on Cantor's theorem on cardinals allows to conclude that the Boolean algebra of finite-cofinite subsets of a countably infinite set doesn't have that shape.
$endgroup$
– amrsa
Jan 16 at 17:48
$begingroup$
It is not true that all Boolean algebras are ${0,1}^n$, for some natural $n$. That is the case for finite Boolean algebras. In general, it isn't even the case that any Boolean algebra is of the type ${0,1}^X$ for some (finite or infinite) set $X$: take, for example the algebra of finite-cofinite subsets of a set. A simple reasoning based on Cantor's theorem on cardinals allows to conclude that the Boolean algebra of finite-cofinite subsets of a countably infinite set doesn't have that shape.
$endgroup$
– amrsa
Jan 16 at 17:48
add a comment |
1 Answer
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Note that $$s=1wedge s=(tvee t')wedge s=(twedge s)vee(t'wedge s).$$ Now $tleq s$ means $twedge s=t$. So, if $t'wedge s$ were $0$, we would get $$s=tvee 0=t$$ which is a contradiction.
answered Jan 16 at 6:58
Eric WofseyEric Wofsey
187k14215344
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$begingroup$
Note that $$s=1wedge s=(tvee t')wedge s=(twedge s)vee(t'wedge s).$$ Now $tleq s$ means $twedge s=t$. So, if $t'wedge s$ were $0$, we would get $$s=tvee 0=t$$ which is a contradiction.
answered Jan 16 at 6:58
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Note that $$s=1wedge s=(tvee t')wedge s=(twedge s)vee(t'wedge s).$$ Now $tleq s$ means $twedge s=t$. So, if $t'wedge s$ were $0$, we would get $$s=tvee 0=t$$ which is a contradiction.
answered Jan 16 at 6:58
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Note that $$s=1wedge s=(tvee t')wedge s=(twedge s)vee(t'wedge s).$$ Now $tleq s$ means $twedge s=t$. So, if $t'wedge s$ were $0$, we would get $$s=tvee 0=t$$ which is a contradiction.
answered Jan 16 at 6:58
Eric WofseyEric Wofsey
187k14215344
$endgroup$
Note that $$s=1wedge s=(tvee t')wedge s=(twedge s)vee(t'wedge s).$$ Now $tleq s$ means $twedge s=t$. So, if $t'wedge s$ were $0$, we would get $$s=tvee 0=t$$ which is a contradiction.
answered Jan 16 at 6:58
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:58
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:58
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:58
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
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It is not true that all Boolean algebras are ${0,1}^n$, for some natural $n$. That is the case for finite Boolean algebras. In general, it isn't even the case that any Boolean algebra is of the type ${0,1}^X$ for some (finite or infinite) set $X$: take, for example the algebra of finite-cofinite subsets of a set. A simple reasoning based on Cantor's theorem on cardinals allows to conclude that the Boolean algebra of finite-cofinite subsets of a countably infinite set doesn't have that shape.
$endgroup$
– amrsa
Jan 16 at 17:48