Computer Science Argumentation Propositional Argumentation

 24 June 14:01   

    Propositional argumentation is a acceptable car to acquaint basal backdrop of logic. It does not accommodate agency to actuate the authority (truth or false) of diminutive statements. Instead, it allows you to appraise the authority of admixture statements accustomed the authority of its diminutive components.

    For example, accede the following:

    :I like Pat or I like Joe.

    :If I like Pat then I like Joe.

    :Do I like Joe?

    Accept as facts the first two statements, acquainted that the use of or actuality is not absolute and appropriately could absolutely be anticipation of as adage I like Pat, or I like Joe, or I like them both. Do these statements betoken that I like Joe is true? Try to argue yourself that I like Joe is true, and accede addition band of reasoning:

    :Pigs can fly or angle can sing.

    :If pigs can fly then angle can sing.

    :Can angle sing?

    We can see that the acknowledgment is yes in both cases. The aloft two sets of statements can be both absent as follows:

    :$P\; vee\; Q$

    :$P$
ightarrow Q

    :$Q$?

    Here, we are anxious about the analytic acumen itself, and not the statements. Thus, instead of alive with fishes or Pats, we artlessly address $Q$s or $P$s. We activate our abstraction first with the syntax of propositional logic: that is, we call the elements in our accent of argumentation and how they are written. We then call the semantics of these symbols : that is, what the symbols mean.

    The syntax of propositional argumentation is composed of propositional symbols , analytic connectives, and parenthesis. Rules administer how these elements can be accounting together. First, we amusement propositional symbols alone as a set of some symbols , for our purposes able-bodied use belletrist of the Roman and Greek alphabets, and accredit to the set of all symbols as $extrm$:

    :Propositional symbols : A set $extrm$ of some symbols . For archetype $p,\; q,\; r,\; ldots$

    Second, we accept the analytic connectives:

    :Logical connectives: $wedge,\; vee,$
eg, o

    Note that these are not the basal appropriate set; they can be analogously represented alone using the individual affiliation NOR (not-or) or NAND (not-and) as is acclimated at the everyman akin in computer hardware.

    Finally, we use parethesis to denote expressions (later on we create departure optional):

    :Parentheses: $(,\; )$

    An announcement is a cord of propositional symbols , parenthesis, and analytic connectives.

    The expressions we accede are alleged formulas. The set $extrm$ of formulas is the aboriginal set of expressions such that:

    #$extrmsubseteq\; extrm$

    #If $phi,\; psi\; in\; extrm$ then

    ##$(phi\; wedge\; psi)in\; extrm$,

    ##$(phi\; vee\; psi)in\; extrm$,

    ##$(phi\; o\; psi)in\; extrm$, and

    ##$($
eg phi)in extrm.

    Another way to ascertain formulas is as the accent authentic by the

    following context-free grammar (with alpha attribute $extrm$):

    :$extrm\; Rightarrow\; extrm$, area $extrm$ stands for any propositional symbol

    :$extrm\; Rightarrow\; (\; extrmwedge\; extrm)$

    :$extrm\; Rightarrow\; (\; extrmvee\; extrm)$

    :$extrm\; Rightarrow\; (\; extrm\; o\; extrm)$

    :$extrm\; Rightarrow\; ($
eg extrm)

    Fact 1 (Unique Readability): The aloft ambience chargeless grammar is unambiguous.

    The action of a blueprint is to make meanings of statements accustomed meanings of diminutive statements. The semantics of a blueprint $phi$ with propositional symbols $p\_1,ldots,\; p\_n$ is a mapping advertence to anniversary accuracy appointment $V$ to $p\_1,ldots,\; p\_n$ a accuracy amount (0 or 1) for $phi$. (The accuracy ethics true and false can be acclimated instead of 1 or 0, respectively, as able-bodied as the abbreviations T and F.)

    The semantics are able-bodied authentic due to Actuality 1.

    One way to specify semantics of a analytic affiliation is via a accuracy table:

    :

    Can one consistently acquisition a blueprint that accouterments any accustomed semantics? Yes, any accuracy table is accomplished by a formula. The blueprint can be begin as follows. Represent the rows area $phi=1$ with conjunctions of the true hypothesis symbols and negations of the false ones. Assuredly address the breach of the results.

    For example,

    :

    $phi:\; (p\; wedge$
eg q)vee (
eg p wedge
eg q)

    Corollary: Every blueprint is agnate to a breach of conjunctions of propositional symbols or antithesis of propositional symbols (DNF).

    Dual of DNF is CNF. To get $phi$ in CNF:

    #Describe cases if $phi$ is false. ex $(pwedge\; q)vee($
eg p wedge q) - DNF $psi$

    #Note that $phi$ is true if $$
eg psi is false. Hence, abate $psi$ using DeMorgans laws.

    ex $($
eg p vee
eg q)wedge (p vee
eg q).

    There are cases if DNF (resp. CNF) is exponentially beyond than the aboriginal formula. For example, for $(x\_1vee\; y\_1)wedge\; (x\_2vee\; y\_2)wedge\; ...wedge\; (x\_nvee\; y\_n)$ the agnate DNF is exponential in size.

    Does anniversary accuracy table accept a polynomial admeasurement blueprint implementing it? Added precisely, does there is $k$ such that every accuracy table with $n$ propositional symbols has a anatomy $phi$ of admeasurement $leq\; n^k$? Answer: no.

    Proof: Accept there exists such $k$. The amount of accuracy tables for $n$ propositional symbols is $2^$.

    The amount of formulas of admeasurement $leq\; n^k$ is $(n+6)^$ ($n$ propositional symbols, 4 connectives and parentheses.)

    Clearly, , for abundantly ample $n$.

    [TODO: account to explain what these definitions are and accommodate their context]

    Note. $phiin\; extrm\; iff$
eg phi in extrm.

    Claim: $Sigma\; models\; phi\; iff\; (Sigma\; cup\; )\; in\; extrm$

    Claim: $extrm$ is NP-complete.

    Proof:

    :$phi\; =\; mathop\_^n\; ((p\_\; wedge$
eg p_ wedge
eg p_)vee (p_ wedge
eg p_ wedge
eg p_)vee (p_ wedge
eg p_ wedge
eg p_))

    :: $wedgemathop\_$
eg (p_ wedge p_)wedge
eg (p_ wedge p_)wedge
eg (p_ wedge p_)

    Claim: $Gin\; extrm\; iff\; phi\; in\; extrm$.

    It is aswell accessible to prove that $extrm\; in\; extrm$ anon

    Claim: $extrm\; in\; extrm$.

    Special case for which SAT is in polynomial time. Example:

    :$(p\; vee$
eg q vee r) wedge (
eg p vee q vee
eg r)

    A Horn article is a breach of literals of which at alotof one is positive. There are two kinds of accessible Horn clauses:

    #clause has 1 absolute literal

    ##$p$, or

    ##$p\; vee$
eg x_1 veeldots vee
eg x_k : x_1 wedge ... wedge x_k
ightarrow p

    #no absolute literal

    ##$$
eg x_1 vee ... vee
eg x_k :
eg (x_1 wedge ... wedge x_k)

    ##$x\_1\; wedge\; ...\; wedge\; x\_k$
ightarrow false

    Claim: For every set $Sigma$ of Horn formulas, blockage whether $Sigma$ is satisfiable is in $extrm$.

    Proof Idea: Let $Sigma\_1$ be the subset of $Sigma$ absolute alone clauses of blazon 1, and $Sigma\_2$ the subset of $Sigma$ absolute clauses of blazon 2. Agenda first that $Sigma\_1$ is satisfiable. To access a minimum acceptable appointment $sigma$, alpha with literals from single-literal clauses and crank the rules. It now charcoal to analysis bendability of $sigma$ with the clauses in $Sigma\_2$. To do this, it is abundant to analysis that for anniversary article $x\_1\; wedge\; ...\; wedge\; x\_k$
ightarrow false in $Sigma\_2$, $sigma$ is not true for all of $x\_1,\; ldots,\; x\_k$.

    Example: Accede the set $Sigma$ of Horn clauses:

    :$p$

    :$q$

    :$r$

    :$$
eg p vee
eg q vee s

    :$$
eg s vee
eg r vee t

    :$$
eg t

    The set $Sigma\_1$ of clauses of blazon 1 consists of the first 5 clauses, and $Sigma\_2$ consists of the endure clause. Agenda that $Sigma\_1$ can aswell be accounting as:

    :$p$

    :$q$

    :$r$

    :$p\; wedge\; q$
ightarrow s

    :$s\; wedge\; r$
ightarrow t

    The minimum acceptable appointment for $Sigma\_1$ is acquired as follows:

    # alpha with $$

    # use the first association to infer $s$

    # use the additional association to infer $t$

    Thus, the minimum acceptable appointment makes $$ true. This contradicts $Sigma\_2$, which states that $t$ haveto be false. Thus, $Sigma$ is not satisfiable.

    A deductive arrangement is a apparatus for proving new statements from accustomed statements.

    Let $Sigma$ be a set of accepted accurate statements (propositional formulas). In a deductive system, there are two components: inference rules and proofs.

    ;Inference rules

    :An inference aphorism indicates that if assertive set of statements (formulas) $varphi\_1,\; ldots,\; varphi\_k$ is true, then a accustomed account $varphi$ haveto be true. An inference aphorism $H$ is denoted as $H:\; frac$.

    :Example (modus ponens): $frac$

    ;Proofs : A affidavit of $varphi$ from $Sigma$ is arrangement of formulas $varphi\_1,\; ...,\; varphi\_n$ such that $varphi\_n=varphi$ and for all $i\; leq\; n$

    ::If $varphi$ has a affidavit from $Sigma$ using inference aphorism $H$ we address $Sigma\; vdash\_H\; varphi$.

    Properties:

    Natural answer is a accumulating of inference rules. Let $perp$ denote contradiction, falsity. The afterward are the inference rules of accustomed deduction:

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\backslash right.$

    # $left\; varphi\; \backslash \; vdots\; \backslash psiend\}$
ight.

    # $left\; varphi\; \backslash \; vdots\; \backslash \; perpend\}$
ight.

    # $left$
egvarphi \ vdots \ perpend}
ight.

    #
ho &
hoend}
ight.

    Rule (13) allows us to prove accurate statements of the anatomy If $varphi$ then $psi$ even if we dont understand the accuracy amount of the $varphi$ account (i.e., $varphi$ is not in the set $Sigma$ of accepted accurate statements). Indeed, for this rule, we alpha bold $varphi$ is valid. If we can achieve $psi$ is accurate in a apple area $Sigmacupvarphi$ are valid, then we achieve that the affiliation $varphi$
ightarrow psi is true, and we absolution the acceptance $varphi$ is valid.

    We now appearance how to administer the aloft inference rules.

    Example: De Morgans Law for negated or-expressions says:

    :$$
eg (varphi vee psi) leftrightarrow (
eg varphi wedge
eg psi)

    Proof: By aphorism $(8)$ if we can prove $$
eg (varphi vee psi)
ightarrow (
eg varphi wedge
eg psi) and $($
eg varphi wedge
eg psi)
ightarrow
eg (varphi vee psi) we can infer the adapted result.

    To prove the first direction, we use aphorism 13 and accept the antecedent $$
eg (varphi vee psi). Then

    :$$
eg (varphi vee psi) (assumed)

    ::$varphi$ (assumed)

    ::$varphi\; vee\; psi$ (by aphorism 13)

    ::$perp$ (by aphorism 5)

    :$$
eg varphi (by aphorism 14)

    ::$psi$ (assumed)

    ::$varphi\; vee\; psi$ (by aphorism 13)

    ::$perp$ (by aphorism 5)

    :$$
eg psi (by aphorism 14)

    :$$
eg varphi wedge
eg psi (by aphorism 1)

    :$$
eg (varphi vee psi)
ightarrow (
eg varphi wedge
eg psi) (by aphorism 13)

    We now prove the additional direction.

    :$$
eg varphi wedge
eg psi (assumed)

    :$$
eg varphi (by aphorism 2)

    :$$
eg psi (by aphorism 3)

    ::$varphi\; vee\; psi$ (assumed)

    :::$varphi\; psi$ (assumed)

    :::$perp\; perp$ (by aphorism 5)

    ::$perp$(by aphorism 16)

    :$$
eg (varphi vee psi) (by aphorism 14)

    :$($
eg varphi wedge
eg psi)
ightarrow
eg (varphi vee psi) (by aphorism 13)

    Proof of Pierces Law:

    :$((A$
ightarrow B)
ightarrow A)
ightarrow A.

    :$(A$
ightarrow B)
ightarrow A (assumed) (1::$$
eg A (assumed)

    :::$A$ (assumed)

    :::$perp$ (by aphorism 5)

    :::$B$ (by aphorism 7)

    ::$A$
ightarrow B (by aphorism 13)

    ::$A$ (by acceptance (1::$perp$ (by aphorism 5)

    :$A$ (by aphorism 14)

    :$(\; (\; A$
ightarrow B )
ightarrow A )
ightarrow A (by aphorism 13)

    Fact 2: Accustomed answer is sound.

    To appearance that accustomed answer is aswell complete we charge to acquaint propositional resolution.

    Resolution is addition action for blockage authority of statements. It involves clauses, formulas and a individual resolution rule.

    Some terminology:

    ;Clause

    :A article is a propositional blueprint composed by breach of literals. For archetype $p\; lor\; q\; lor\; lnot\; r$. It is usually denoted as the set of literals, e.g. $$.

    :The abandoned clause, denoted as an accessible box $Box$, is the breach of no literals. It is consistently false.

    ;Formula

    :A set of clauses, anniversary of them satisfiable. For example, $,,\backslash \}$ represents the CNF blueprint $(p\; lorlnot\; q)land\; (r)land\; (lnot\; r\; lor\; s)$.

    :The abandoned formula, denoted as $emptyset$, is the set that contains no clauses. It is consistently true.

    ;Resolution Aphorism

    :It is a aphorism that, accustomed two clauses $C$ (containing some accurate $y$) and $C$ (containing some accurate $lnot\; y$), allows to infer a new clause, alleged the resolvent of $C$ and $C$ (with account to $y$).

    A affidavit arrangement for resolution contains a individual resolution rule, area the resolvent is authentic as follows. Accept $C$ and $C$ are clauses such that $yin\; C$ and $lnot\; yin\; C$, then

    :$res\_(C,C)\; =\; (C-)cup\; (C-)$.

    The aboriginal set of clauses absolute $varphi$ and bankrupt beneath resolution is denoted $Res(varphi)$.

    Example: If $C=$ and $C=$, then $res\_y(C,C)\; =$.

    It is accessible to appearance that the resolution rule, as defined, computes a article that can be accepted using accustomed deduction.

    Claim: Let $C$ and $C$ be any two clauses such that $yin\; C$ and $lnot\; yin\; C$. Then $Cland\; C\; implies\; res\_y(C,C)$.

    In adjustment to prove the authority of a account $psi$, we will prove the negated account $lnot\; psi$ is unsatisfiable. To prove unsatisfiability of a blueprint $varphi$, we charge to ascertain the resolution acknowledgment of the blueprint $varphi$:

    The resolution acknowledgment timberline of the blueprint $varphi$ is a timberline abiding at the abandoned clause, area every blade is a article in $varphi$ and anniversary centralized bulge is computed as the resolvent of the two agnate children.

    Notice that clauses of $varphi$ can arise again as leaves. From aloft affirmation we can achieve that:

    Claim: If there exists a resolution acknowledgment timberline for blueprint $varphi$, then $varphi\; implies\; Box$, that is, $varphi$ is unsatisfiable.

    Example: The formula

    :$varphi\; =\; (plor\; q)land\; (lnot\; qlor\; r)land\; (lnot\; r)land\; (lnot\; p\; lor\; lnot\; s)land\; (slor\; lnot\; t)\; land\; (t)$

    has the afterward resolution acknowledgment tree:

    

    The adjustment in which clauses are called to compute the resolvent affairs if accretion the resolution acknowledgment tree, as the afterward archetype shows: Accede the formula

    :$psi\; =\; (plor\; q)land\; (lnot\; qlor\; r)land\; (lnot\; p)land\; (lnot\; q)$.

    Even admitting a resolution acknowledgment timberline may is for $psi$, adjustment is important if aggravating to body the tree. Beneath are two altered resolution acknowledgment trees, but alone one is successful:

    Soundness: Propositional resolution is sound, that is, if there exists a resolution acknowledgment timberline for a accustomed blueprint $varphi$, then $varphi$ haveto be unsatisfiable.

    Theorem: For any blueprint $varphi$, if $Box\; in\; Res(varphi)$, then $varphi\; implies\; Box$.

    Completeness: Propositional resolution is complete, that is, if a accustomed blueprint $varphi$ is unsatisfiable, then $varphi$ has a resolution acknowledgment tree.

    Theorem: For any blueprint $varphi$, if $varphi\; implies\; Box$, then $Box\; in\; Res(varphi)$.

    Proof: By consecration on the amount of variables in $varphi$.

    Basis: We accept one variable, say $p$. All accessible clauses of $varphi$ are $$ and $$. If $varphi$ is clamorous then both clauses occur, and accordingly $Box\; in\; Res(varphi)$.

    Induction step: Accept the antecedent is true for formulas with beneath than $n$ variables. Let $varphi$ be a blueprint with $n$ variables. Accept $Box$
otin Res(varphi); we will appearance $varphi$ is satisfiable. Let $p$ be a capricious of $varphi$. Then either $\backslash notin\; Res(varphi)$ or $\backslash notin\; Res(varphi)$ (if both authority then $Boxin\; Res(varphi)$ immediately).

    Assume $\backslash notin\; Res(varphi)$. We ascertain the blueprint $varphi^p$ as absolute all clauses that do not accommodate $$ and area the accurate $lnot\; p$ has been removed from anniversary article (in additional words, $varphi^p$ is agnate to the blueprint consistent from ambience $p$ true).

    Formally,

    :$varphi^p\; =$.

    First, apprehension that

    :$Res(varphi^p)\; =$

    and thus,

    :$\backslash notin\; Res(varphi^p)$.

    Also, back $Box$
otin Res(varphi) we accept that $Box$
otin Res(varphi^p). By the consecration hypothesis, $varphi^p$ is satisfiable. Then $varphi$ is satisfiable by an addendum of the acceptable appointment of $varphi^p$ with $p$ according true. The case $\backslash in\; Res(varphi)$ is analogous.

    Theorem: Let $H$ be the set of inference rules of Accustomed Deduction. If $Sigma\; models\; varphi$ then $Sigma\; vdash\_H\; varphi$.

    The abstraction abaft the affidavit of abyss of accustomed answer is as follows. Accept $varphi$ is accurate (then $lnot\; varphi$ is unsatisfiable). We then appearance there exists a resolution acknowledgment for $varphi$ and then by applying the bucking aphorism (rule 15):

    :$frac$

    we achieve $varphi$ can be inferred.

    Proof: (Sketch) Accustomed a blueprint $varphi$ accurate beneath $Sigma$, we accomplish the afterward steps:

    #Prove that $lnot\; varphi$ is agnate to some $psi$, area $psi$ is in CNF.

    #Prove that $psi\; implies\; Res(psi)$, for all $psi$.

    #By abyss of resolution, if $psi$ is clamorous then $Box\; in\; Res(psi)$. Therefore, $$ and $in\; Res(psi)$ for some accurate $p$. This implies .

    #Conclude that and accordingly $varphi$ is valid.

    Step (1) can be calmly done by again appliance of De Morgans laws. Move (2) can be accurate using accustomed deduction. Finally, move (3) can be accurate by consecration on the amount of accomplish to access $Res(psi)$. Clearly, anniversary move can be apish using accustomed deduction.

    It is actual acceptable that any algorithm for propositional resolution will yield actual continued on the affliction case (recall that blockage authority of a blueprint $varphi$ is co-NP complete).

    Linear resolution is a accurate resolution action that consistently resolves the alotof contempo resolvent with a clause. The resolution acknowledgment timberline so acquired is accordingly linear. It is accessible to prove that, if the set of clauses are Horn clauses, there exists a beeline resolution action for any formula. That is, beeline resolution is complete for the set of Horn clauses.

    The accent PROLOG uses resolution on a set of Horn clauses. Anniversary article is alleged a program clause. Moreover, clauses composed by a individual accurate are alleged facts.

    A article with a individual negated accurate is alleged a query. The table beneath shows a allegory of the altered notations. In PROLOG, to concern a account $t$, the abstraction is to abate the account ($lnot\; t$) and to accomplish resolution with the set of accepted true statements. If a resolution acknowledgment timberline is found, the account $t$ is adumbrated by the program.

    Example: An archetype of beeline resolution for the formula

    :$phi\; =\; (p)land\; (q)land\; (r)land\; (tlor\; lnot\; s\; lor\; lnot\; r)land\; (s\; lor\; lnot\; p\; lor\; lnot\; q)land\; (lnot\; t)$

    is apparent here: