Verifying a proof regarding the duals of two equivalent compound propositions also being equivalent












1














I am aware that there are many previously asked questions regarding the proof that the duals of two equivalent compound propositions containing only logical operators ∨, ∧, and ¬ are also equivalent, although I have not yet been content with the solutions I have found.



If possible, I would like someone qualified to see if I am on the right track with my current proof of the above. Please note that I am only at a relatively basic level for proof-writing, although I wish to advance this whenever possible.



Below is a definition of the dual of a compound proposition containing only logical operators ∨, ∧, and ¬, found in Discrete Mathematics and Its Applications, 7th Edition, by Rosen, followed by the relevant theorem and current draft of my proof.




The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s∗.




 




Theorem:

The duals of two equivalent compound propositions are also equivalent.



Proof:

Let $s_0$, $s_n$ be two arbitrary compound propositions containing
only logical operators ∨, ∧, or ¬, such that $s_0$ $equiv$ $s_n$
after transforming $s_0$ with n logical equivalences containing
only logical operators ∨, ∧, or ¬.



Let $t_0$, $t_1$, $ldots$ , tn-1 be an ordered sequence of
n transformations using logical equivalences only containing the ∨, ∧,
or ¬ operators such that $t_i$($s_i$) = si+1, for 0 $leq$
i $leq$ n-1, where $s_i$, si+1 are compound propositions, and let
$s_0^*$, $s_1^*$, $ldots$ , $s_n^*$ be the
corresponding duals for $s_0$, $s_1$, $ldots$ , $s_n$, respectively.



We wish to find an ordered list of m transformations, $t_0^*$, $t_1^*$,
$ldots$ , tm-1$^*$ such that $t_i^*$($s_i^*$) =
si+1$^*$,
for 0 $leq$ i $leq$ m-1, where $s_i$,
si+1 are compound propositions.



Since every logical equivalence for a compound proposition containing only
the ∨, ∧, or ¬ operators has a dual equivalent, that is, a complementary
logical equivalence for the dual of the compound proposition, we let the
logical equivalence transformation $t_i^*$ be the dual equivalent of the
transformation $t_i$ when the transformation is performed on a compound
proposition, and otherwise the same transformation when performed on an
individual element,
for 0 $leq$ i $leq$ n-1.



Thus, $t_i^*$($s_i^*$) = si+1$^*$, for 0 $leq$ i $leq$ n-1,
is a valid ordered sequence of logical equivalence transformations, and $s_0^*$
$equiv$ $s_n^*$. Since $s_0^*$ and $s_n^*$ were chosen to be arbitrary,
true for all such.



End of proof.




I feel that the first statement in last paragraph of my proof is not explicit enough, although I am not currently able to think of what is additionally needed.



Please help if possible. Thank you!










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  • Yes, this is correct.
    – Berci
    Dec 26 '18 at 22:53
















1














I am aware that there are many previously asked questions regarding the proof that the duals of two equivalent compound propositions containing only logical operators ∨, ∧, and ¬ are also equivalent, although I have not yet been content with the solutions I have found.



If possible, I would like someone qualified to see if I am on the right track with my current proof of the above. Please note that I am only at a relatively basic level for proof-writing, although I wish to advance this whenever possible.



Below is a definition of the dual of a compound proposition containing only logical operators ∨, ∧, and ¬, found in Discrete Mathematics and Its Applications, 7th Edition, by Rosen, followed by the relevant theorem and current draft of my proof.




The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s∗.




 




Theorem:

The duals of two equivalent compound propositions are also equivalent.



Proof:

Let $s_0$, $s_n$ be two arbitrary compound propositions containing
only logical operators ∨, ∧, or ¬, such that $s_0$ $equiv$ $s_n$
after transforming $s_0$ with n logical equivalences containing
only logical operators ∨, ∧, or ¬.



Let $t_0$, $t_1$, $ldots$ , tn-1 be an ordered sequence of
n transformations using logical equivalences only containing the ∨, ∧,
or ¬ operators such that $t_i$($s_i$) = si+1, for 0 $leq$
i $leq$ n-1, where $s_i$, si+1 are compound propositions, and let
$s_0^*$, $s_1^*$, $ldots$ , $s_n^*$ be the
corresponding duals for $s_0$, $s_1$, $ldots$ , $s_n$, respectively.



We wish to find an ordered list of m transformations, $t_0^*$, $t_1^*$,
$ldots$ , tm-1$^*$ such that $t_i^*$($s_i^*$) =
si+1$^*$,
for 0 $leq$ i $leq$ m-1, where $s_i$,
si+1 are compound propositions.



Since every logical equivalence for a compound proposition containing only
the ∨, ∧, or ¬ operators has a dual equivalent, that is, a complementary
logical equivalence for the dual of the compound proposition, we let the
logical equivalence transformation $t_i^*$ be the dual equivalent of the
transformation $t_i$ when the transformation is performed on a compound
proposition, and otherwise the same transformation when performed on an
individual element,
for 0 $leq$ i $leq$ n-1.



Thus, $t_i^*$($s_i^*$) = si+1$^*$, for 0 $leq$ i $leq$ n-1,
is a valid ordered sequence of logical equivalence transformations, and $s_0^*$
$equiv$ $s_n^*$. Since $s_0^*$ and $s_n^*$ were chosen to be arbitrary,
true for all such.



End of proof.




I feel that the first statement in last paragraph of my proof is not explicit enough, although I am not currently able to think of what is additionally needed.



Please help if possible. Thank you!










share|cite|improve this question









New contributor




christophercrary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Yes, this is correct.
    – Berci
    Dec 26 '18 at 22:53














1












1








1







I am aware that there are many previously asked questions regarding the proof that the duals of two equivalent compound propositions containing only logical operators ∨, ∧, and ¬ are also equivalent, although I have not yet been content with the solutions I have found.



If possible, I would like someone qualified to see if I am on the right track with my current proof of the above. Please note that I am only at a relatively basic level for proof-writing, although I wish to advance this whenever possible.



Below is a definition of the dual of a compound proposition containing only logical operators ∨, ∧, and ¬, found in Discrete Mathematics and Its Applications, 7th Edition, by Rosen, followed by the relevant theorem and current draft of my proof.




The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s∗.




 




Theorem:

The duals of two equivalent compound propositions are also equivalent.



Proof:

Let $s_0$, $s_n$ be two arbitrary compound propositions containing
only logical operators ∨, ∧, or ¬, such that $s_0$ $equiv$ $s_n$
after transforming $s_0$ with n logical equivalences containing
only logical operators ∨, ∧, or ¬.



Let $t_0$, $t_1$, $ldots$ , tn-1 be an ordered sequence of
n transformations using logical equivalences only containing the ∨, ∧,
or ¬ operators such that $t_i$($s_i$) = si+1, for 0 $leq$
i $leq$ n-1, where $s_i$, si+1 are compound propositions, and let
$s_0^*$, $s_1^*$, $ldots$ , $s_n^*$ be the
corresponding duals for $s_0$, $s_1$, $ldots$ , $s_n$, respectively.



We wish to find an ordered list of m transformations, $t_0^*$, $t_1^*$,
$ldots$ , tm-1$^*$ such that $t_i^*$($s_i^*$) =
si+1$^*$,
for 0 $leq$ i $leq$ m-1, where $s_i$,
si+1 are compound propositions.



Since every logical equivalence for a compound proposition containing only
the ∨, ∧, or ¬ operators has a dual equivalent, that is, a complementary
logical equivalence for the dual of the compound proposition, we let the
logical equivalence transformation $t_i^*$ be the dual equivalent of the
transformation $t_i$ when the transformation is performed on a compound
proposition, and otherwise the same transformation when performed on an
individual element,
for 0 $leq$ i $leq$ n-1.



Thus, $t_i^*$($s_i^*$) = si+1$^*$, for 0 $leq$ i $leq$ n-1,
is a valid ordered sequence of logical equivalence transformations, and $s_0^*$
$equiv$ $s_n^*$. Since $s_0^*$ and $s_n^*$ were chosen to be arbitrary,
true for all such.



End of proof.




I feel that the first statement in last paragraph of my proof is not explicit enough, although I am not currently able to think of what is additionally needed.



Please help if possible. Thank you!










share|cite|improve this question









New contributor




christophercrary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I am aware that there are many previously asked questions regarding the proof that the duals of two equivalent compound propositions containing only logical operators ∨, ∧, and ¬ are also equivalent, although I have not yet been content with the solutions I have found.



If possible, I would like someone qualified to see if I am on the right track with my current proof of the above. Please note that I am only at a relatively basic level for proof-writing, although I wish to advance this whenever possible.



Below is a definition of the dual of a compound proposition containing only logical operators ∨, ∧, and ¬, found in Discrete Mathematics and Its Applications, 7th Edition, by Rosen, followed by the relevant theorem and current draft of my proof.




The dual of a compound proposition that contains only the
logical operators ∨, ∧, and ¬ is the compound proposition
obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s∗.




 




Theorem:

The duals of two equivalent compound propositions are also equivalent.



Proof:

Let $s_0$, $s_n$ be two arbitrary compound propositions containing
only logical operators ∨, ∧, or ¬, such that $s_0$ $equiv$ $s_n$
after transforming $s_0$ with n logical equivalences containing
only logical operators ∨, ∧, or ¬.



Let $t_0$, $t_1$, $ldots$ , tn-1 be an ordered sequence of
n transformations using logical equivalences only containing the ∨, ∧,
or ¬ operators such that $t_i$($s_i$) = si+1, for 0 $leq$
i $leq$ n-1, where $s_i$, si+1 are compound propositions, and let
$s_0^*$, $s_1^*$, $ldots$ , $s_n^*$ be the
corresponding duals for $s_0$, $s_1$, $ldots$ , $s_n$, respectively.



We wish to find an ordered list of m transformations, $t_0^*$, $t_1^*$,
$ldots$ , tm-1$^*$ such that $t_i^*$($s_i^*$) =
si+1$^*$,
for 0 $leq$ i $leq$ m-1, where $s_i$,
si+1 are compound propositions.



Since every logical equivalence for a compound proposition containing only
the ∨, ∧, or ¬ operators has a dual equivalent, that is, a complementary
logical equivalence for the dual of the compound proposition, we let the
logical equivalence transformation $t_i^*$ be the dual equivalent of the
transformation $t_i$ when the transformation is performed on a compound
proposition, and otherwise the same transformation when performed on an
individual element,
for 0 $leq$ i $leq$ n-1.



Thus, $t_i^*$($s_i^*$) = si+1$^*$, for 0 $leq$ i $leq$ n-1,
is a valid ordered sequence of logical equivalence transformations, and $s_0^*$
$equiv$ $s_n^*$. Since $s_0^*$ and $s_n^*$ were chosen to be arbitrary,
true for all such.



End of proof.




I feel that the first statement in last paragraph of my proof is not explicit enough, although I am not currently able to think of what is additionally needed.



Please help if possible. Thank you!







discrete-mathematics proof-verification logic proof-writing






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christophercrary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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Check out our Code of Conduct.









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edited Dec 26 '18 at 22:45





















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asked Dec 26 '18 at 22:36









christophercrary

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christophercrary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • Yes, this is correct.
    – Berci
    Dec 26 '18 at 22:53


















  • Yes, this is correct.
    – Berci
    Dec 26 '18 at 22:53
















Yes, this is correct.
– Berci
Dec 26 '18 at 22:53




Yes, this is correct.
– Berci
Dec 26 '18 at 22:53















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