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Propositional Logic: introduction to this part

This part contains material on classical propositional logic. The first chapter is relatively rudimentary and just lists definitions and results, many proofs are not carried out but are left as exercises. The material on proof systems and the completeness theorem is included from the part on first-order logic, with the “FOL” tag set to false. This leaves out everything related to predicates, terms, and quantifiers, and replaces talk of structures M\Struct{M}source 17 with talk about valuations v\pAssign{v}source 18.

It is planned to expand this part to include more detail, and to add further topics and results, such as truth-functional completeness.

Introduction

Propositional logic deals with formulas that are built from propositional variables using the propositional connectives ¬\lnotsource 15, \landsource 15, \lorsource 15, \lifsource 15, and \liffsource 15. Intuitively, a propositional variable ppsource 16 stands for a sentence or proposition that is true or false. Whenever the “truth value” of the propositional variable in a formula is determined, so is the truth value of any formulas formed from them using propositional connectives. We say that propositional logic is truth functional, because its semantics is given by functions of truth values. In particular, in propositional logic we leave out of consideration any further determination of truth and falsity, e.g., whether something is necessarily true rather than just contingently true, or whether something is known to be true, or whether something is true now rather than was true or will be true. We only consider two truth values true (True\Truesource 27) and false (False\Falsesource 27), and so exclude from discussion the possibility that a statement may be neither true nor false, or only half true. We also concentrate only on connectives where the truth value of a formula built from them is completely determined by the truth values of its parts (and not, say, on its meaning). In particular, whether the truth value of conditionals in English is truth functional in this sense is contentious. The material conditional \lifsource 34 is; other logics deal with conditionals that are not truth functional.

In order to develop the theory and metatheory of truth-functional propositional logic, we must first define the syntax and semantics of its expressions. We will describe one way of constructing formulas from propositional variables using the connectives. Alternative definitions are possible. Other systems will choose different symbols, will select different sets of connectives as primitive, and will use parentheses differently (or even not at all, as in the case of so-called Polish notation). What all approaches have in common, though, is that the formation rules define the set of formulas inductively. If done properly, every expression can result essentially in only one way according to the formation rules. The inductive definition resulting in expressions that are uniquely readable means we can give meanings to these expressions using the same method—inductive definition.

Giving the meaning of expressions is the domain of semantics. The central concept in semantics for propositional logic is that of satisfaction in a valuation. A valuation v\pAssign{v}source 54 assigns truth values True\Truesource 55, False\Falsesource 55 to the propositional variables. Any valuation determines a truth value v¯(A)\pValue{v}(!A)source 56 for any formula A!Asource 57. A formula is satisfied in a valuation v\pAssign{v}source 58 iff v¯(A)=True\pValue{v}(!A) = \Truesource 58—we write this as vA\pSat{v}{!A}source 59. This relation can also be defined by induction on the structure of A!Asource 60, using the truth functions for the logical connectives to define, say, satisfaction of AB!A \land !Bsource 61 in terms of satisfaction (or not) of A!Asource 62 and B!Bsource 62.

On the basis of the satisfaction relation vA\pSat{v}{!A}source 64 for sentences we can then define the basic semantic notions of tautology, entailment, and satisfiability. A formula is a tautology, A\Entails !Asource 67, if every valuation satisfies it, i.e., v¯(A)=True\pValue{v}(!A) = \Truesource 68 for any v\pAssign{v}source 68. It is entailed by a set of formulas, ΓA\Gamma \Entails !Asource 69, if every valuation that satisfies all the formulas in Γ\Gammasource 70 also satisfies A!Asource 70. And a set of formulas is satisfiable if some valuation satisfies all formulas in it at the same time. Because formulas are inductively defined, and satisfaction is in turn defined by induction on the structure of formulas, we can use induction to prove properties of our semantics and to relate the semantic notions defined.

Propositional formula

Formulas of propositional logic are built up from propositional variables and the propositional constant \lfalsesource 16 using logical connectives.

  1. A denumerable set Prop\PVarsource 21 of propositional variables p0\Obj p_0source 21, p1\Obj p_1source 22, … The propositional constant for falsity \lfalsesource 23.

  2. The logical connectives: , ¬\lnotsource 27 (negation) , \landsource 28 (conjunction) , \lorsource 29 (disjunction) , \lifsource 30 (conditional)

  3. Punctuation marks: (, ), and the comma.

We denote this language of propositional logic by L0\Lang L_0source 35.

In addition to the primitive connectives introduced above, we also use the following defined symbols:

, \liffsource 45 (biconditional)

, \ltruesource 47 (truth).

Definition: Propositional formulas

The set FrmL0\Frm[L_0]source 80 of formulas of propositional logic is defined inductively as follows:

\lfalsesource 83 is an atomic formula.

  1. Every propositional variable pi\Obj p_isource 87 is an atomic formula.

    If A!Asource 90 is a formula, then ¬A\lnot !Asource 90 is a formula.

    If A!Asource 93 and B!Bsource 93 are formulas, then (AB)(!A \land !B)source 93 is a formula.

    If A!Asource 96 and B!Bsource 96 are formulas, then (AB)(!A \lor !B)source 96 is a formula.

    If A!Asource 99 and B!Bsource 99 are formulas, then (AB)(!A \lif !B)source 99 is a formula.

    Nothing else is a formula.

source 78

When writing a formula (BC)(!B \ast !C)source 129 constructed from B!Bsource 129, C!Csource 129 using a two-place connective \astsource 130, we will often leave out the outermost pair of parentheses and write simply BC!B \ast !Csource 131.

Definition: Defined propositional operators

Formulas constructed using the defined operators are to be understood as follows:

\ltruesource 140 abbreviates ¬\lnot\lfalsesource 141.

AB!A \liff !Bsource 161 abbreviates (AB)(BA)(!A \lif !B) \land (!B \lif !A)source 161.

    source 134

    Definition: Syntactic identity

    The symbol \identsource 168 expresses syntactic identity between strings of symbols, i.e., AB!A \ident !Bsource 169 iff A!Asource 169 and B!Bsource 169 are strings of symbols of the same length and which contain the same symbol in each place.

    source 167

    The \identsource 173 symbol may be flanked by strings obtained by concatenation, e.g., A(BC)!A \ident (!B \lor !C)source 174 means: the string of symbols A!Asource 175 is the same string as the one obtained by concatenating an opening parenthesis, the string B!Bsource 176, the \lorsource 176 symbol, the string C!Csource 177, and a closing parenthesis, in this order. If this is the case, then we know that the first symbol of A!Asource 178 is an opening parenthesis, A!Asource 179 contains B!Bsource 179 as a substring (starting at the second symbol), that substring is followed by \lorsource 180, etc.

    Preliminaries

    Theorem: Structural induction for formulas

    If some property PPsource 15 holds for all the atomic formulas and is such that

    it holds for ¬A\lnot !Asource 18 whenever it holds for A!Asource 19; it holds for (AB)(!A \land !B)source 20 whenever it holds for A!Asource 21 and B!Bsource 21; it holds for (AB)(!A \lor !B)source 22 whenever it holds for A!Asource 23 and B!Bsource 23; it holds for (AB)(!A \lif !B)source 24 whenever it holds for A!Asource 25 and B!Bsource 25;

      then PPsource 29 holds for all formulas.

      source 13

      Proof

      Let SSsource 33 be the collection of all formulas with property PPsource 34. Clearly SFrmL0S \subseteq \Frm[L_0]source 34. SSsource 34 satisfies all the conditions of Definition: Propositional formulas: it contains all atomic formulas and is closed under the operators. FrmL0\Frm[L_0]source 36 is the smallest such class, so FrmL0S\Frm[L_0] \subseteq Ssource 37. So FrmL0=S\Frm[L_0] = Ssource 37, and every formula has property PPsource 38.

      End of proof.

      Proposition: Balanced parentheses in formulas

      Any formula in FrmL0\Frm[L_0]source 42 is balanced, in that it has as many left parentheses as right ones.

      source 41

      Exercise: Prove balanced parentheses

      Prove Proposition: Balanced parentheses in formulas

      source 46

      Proposition: No proper initial segment is a formula

      No proper initial segment of a formula is a formula.

      source 50

      Exercise: Prove the initial-segment proposition

      Prove Proposition: No proper initial segment is a formula

      source 54

      Proposition: Unique readability of formulas

      Any formula A!Asource 59 in FrmL0\Frm[L_0]source 59 has exactly one parsing as one of the following

      \lfalsesource 62.

      1. pn\Obj p_nsource 66 for some pnProp\Obj p_n \in \PVarsource 66.

        ¬B\lnot !Bsource 68 for some formula B!Bsource 68.

        (BC)(!B \land !C)source 70 for some formulas B!Bsource 70 and C!Csource 70.

        (BC)(!B \lor !C)source 72 for some formulas B!Bsource 72 and C!Csource 72.

        (BC)(!B \lif !C)source 74 for some formulas B!Bsource 74 and C!Csource 74.

      Moreover, this parsing is unique.

      source 58

      Proof

      By induction on A!Asource 82. For instance, suppose that A!Asource 82 has two distinct readings as (BC)(!B \lif !C)source 83 and (B'C')(!B' \lif !C')source 83. Then B!Bsource 83 and B'!B'source 83 must be the same (or else one would be a proper initial segment of the other); so if the two readings of A!Asource 85 are distinct it must be because C!Csource 86 and C'!C'source 86 are distinct readings of the same sequence of symbols, which is impossible by the inductive hypothesis.

      End of proof.

      Definition: Uniform substitution

      If A!Asource 92 and B!Bsource 92 are formulas, and pi\Obj p_isource 92 is a propositional variable, then A[B/pi]\Subst{!A}{!B}{\Obj p_i}source 93 denotes the result of replacing each occurrence of pi\Obj p_isource 94 by an occurrence of B!Bsource 94 in A!Asource 94; similarly, the simultaneous substitution of p1\Obj p_1source 95, …, pn\Obj p_nsource 95 by formulas B1!B_1source 96, …, Bn!B_nsource 96 is denoted by A[[B1/p1],,[Bn/pn]]\SSubst{!A}{\subst{!B_1}{\Obj p_1},\dots,\subst{!B_n}{\Obj p_n}}source 97.

      source 91

      Exercise: Identify uniform substitutions

      For each of the five formulas below determine whether the formula can be expressed as a substitution A[B/pi]\Subst{!A}{!B}{\Obj p_i}source 101 where A!Asource 102 is (i) p0\Obj p_0source 102; (ii) (¬p0p1)( \lnot \Obj p_0 \land \Obj p_1)source 102; and (iii) ((¬p0p1)p2)( ( \lnot \Obj p_0 \lif \Obj p_1 ) \land \Obj p_2 )source 103. In each case specify the relevant substitution.

      1. p1\Obj p_1source 106

      2. (¬p0p0)( \lnot \Obj p_0 \land \Obj p_0 )source 107

      3. ((p0p1)p2)( ( \Obj p_0 \lor \Obj p_1 ) \land \Obj p_2 )source 108

      4. ¬((p0p1)p2)\lnot ( ( \Obj p_0 \lif \Obj p_1 ) \land \Obj p_2 )source 109

      5. ((¬(p0p1)(p0p1))¬(p0p1))(( \lnot ( \Obj p_0 \lif \Obj p_1 ) \lif ( \Obj p_0 \lor \Obj p_1 )) \land \lnot ( \Obj p_0 \land \Obj p_1 ))source 110

      source 100

      Exercise: Define substitution by induction

      Give a mathematically rigorous definition of A[B/p]\Subst{!A}{!B}{p}source 115 by induction.

      source 114

      Formation Sequences

      Defining formulas via an inductive definition, and the complementary technique of proving properties of formulas via induction, is an elegant and efficient approach. However, it can also be useful to consider a more bottom-up, step-by-step approach to the construction of formulas, which we do here using the notion of a formation sequence.

      Definition: Formation sequences for formulas

      A finite sequence A0,,An\tuple{!A_0,\dotsc,!A_n}source 22 of strings of symbols from the language L0\Lang L_0source 23 is a formation sequence for A!Asource 24 if AAn!A \ident !A_nsource 24 and for all ini \leq nsource 24, either Ai!A_isource 25 is an atomic formula or there exist j,k<ij,k < isource 25 such that one of the following holds:

      Ai¬Aj!A_i \ident \lnot !A_jsource 28. Ai(AjAk)!A_i \ident (!A_j \land !A_k)source 29. Ai(AjAk)!A_i \ident (!A_j \lor !A_k)source 30. Ai(AjAk)!A_i \ident (!A_j \lif !A_k)source 31.

        source 20

        Example: A formation sequence

        p0,p1,(p1p0),¬(p1p0)\tuple{ \Obj p_0, \Obj p_1, (\Obj p_1 \land \Obj p_0), \lnot (\Obj p_1 \land \Obj p_0) }source 37 is a formation sequence of ¬(p1p0)\lnot (\Obj p_1 \land \Obj p_0)source 46, as is p0,p1,p0,(p1p0),(p0p1),¬(p1p0).\tuple{ \Obj p_0, \Obj p_1, \Obj p_0, (\Obj p_1 \land \Obj p_0), (\Obj p_0 \lif \Obj p_1), \lnot (\Obj p_1 \land \Obj p_0) }.source 47

        As can be seen from the second example, formation sequences may contain `junk': formulas which are redundant or do not contribute to the construction.

        source 36

        Proposition: Every formula has a formation sequence

        Every formula A!Asource 65 in FrmL0\Frm[L_0]source 65 has a formation sequence.

        source 63

        Proof

        Suppose A!Asource 69 is atomic. Then the sequence A\tuple{!A}source 69 is a formation sequence for A!Asource 70.

        Now suppose that B!Bsource 72 and C!Csource 72 have formation sequences B0,,Bn\tuple{!B_0,\dotsc,!B_n}source 73 and C0,,Cm\tuple{!C_0,\dotsc,!C_m}source 73 respectively.

        If A¬B!A \ident \lnot !Bsource 77, then B0,,Bn,¬Bn\tuple{!B_0,\dotsc,!B_n,\lnot !B_n}source 78 is a formation sequence for A!Asource 79. If A(BC)!A \ident (!B \land !C)source 80, then B0,,Bn,C0,,Cm,(BnCm)\tuple{!B_0,\dotsc,!B_n,!C_0,\dotsc,!C_m,(!B_n \land !C_m)}source 81 is a formation sequence for A!Asource 82. If A(BC)!A \ident (!B \lor !C)source 83, then B0,,Bn,C0,,Cm,(BnCm)\tuple{!B_0,\dotsc,!B_n,!C_0,\dotsc,!C_m,(!B_n \lor !C_m)}source 84 is a formation sequence for A!Asource 85. If A(BC)!A \ident (!B \lif !C)source 86, then B0,,Bn,C0,,Cm,(BnCm)\tuple{!B_0,\dotsc,!B_n,!C_0,\dotsc,!C_m,(!B_n \lif !C_m)}source 87 is a formation sequence for A!Asource 88.

          By the principle of induction on formulas, every formula has a formation sequence.

          End of proof.

          We can also prove the converse. This is important because it shows that our two ways of defining formulas are equivalent: they give the same results. It also means that we can prove theorems about formulas by using ordinary induction on the length of formation sequences.

          Lemma: Initial subsequences remain formation sequences

          Suppose that A0,,An\tuple{!A_0,\dotsc,!A_n}source 105 is a formation sequence for An!A_nsource 106, and that knk \leq nsource 106. Then A0,,Ak\tuple{!A_0,\dotsc,!A_k}source 106 is a formation sequence for Ak!A_ksource 107.

          source 103

          Proof

          Exercise.

          End of proof.

          Theorem: Formulas characterized by formation sequences

          FrmL0\Frm[L_0]source 116 is the set of all strings of symbols in the language L0\Lang L_0source 117 with a formation sequence.

          source 114

          Proof

          Let FFsource 121 be the set of all strings of symbols in the language L0\Lang L_0source 122 that have a formation sequence. We have seen in Proposition: Every formula has a formation sequence that FrmL0F\Frm[L_0] \subseteq Fsource 124, so now we prove the converse.

          Suppose A!Asource 126 has a formation sequence A0,,An\tuple{!A_0,\dotsc,!A_n}source 126. We prove that AFrmL0!A \in \Frm[L_0]source 127 by strong induction on nnsource 127. Our induction hypothesis is that every string of symbols with a formation sequence of length m<nm < nsource 129 is in FrmL0\Frm[L_0]source 129. By the definition of a formation sequence, either An!A_nsource 130 is atomic or there must exist j,k<nj,k < nsource 131 such that one of the following is the case:

          An¬Aj!A_n \ident \lnot !A_jsource 134. An(AjAk)!A_n \ident (!A_j \land !A_k)source 135. An(AjAk)!A_n \ident (!A_j \lor !A_k)source 136. An(AjAk)!A_n \ident (!A_j \lif !A_k)source 137.

            Now we reason by cases. If An!A_nsource 140 is atomic then AnFrmL0!A_n \in \Frm[L_0]source 141. Suppose instead that A(AjAk)!A \equiv (!A_j \land !A_k)source 141. By Lemma: Initial subsequences remain formation sequences, A0,,Aj\tuple{!A_0,\dotsc,!A_j}source 143 and A0,,Ak\tuple{!A_0,\dotsc,!A_k}source 143 are formation sequences for Aj!A_jsource 144 and Ak!A_ksource 144 respectively. Since these are proper initial subsequences of the formation sequence for A!Asource 146, they both have length less than nnsource 146. Therefore by the induction hypothesis, Aj!A_jsource 147 and Ak!A_ksource 147 are in FrmL0\Frm[L_0]source 147, and so by the definition of a formula, so is (AjAk)(!A_j \land !A_k)source 149. The other cases follow by parallel reasoning.

            End of proof.

            valuation and Satisfaction

            Definition: Propositional valuations

            Let {True,False}\{\True, \False\}source 14 be the set of the two truth values, “true” and “false.” A valuation for L0\Lang{L_0}source 15 is a function v\pAssign{v}source 16 assigning either True\Truesource 16 or False\Falsesource 16 to the propositional variables of the language, i.e., v:Prop{True,False}\pAssign{v} \colon \PVar \to \{\True, \False \}source 17.

            source 13

            Definition: Evaluation of propositional formulas

            Given a valuation v\pAssign{v}source 22, define the evaluation function v¯:FrmL0{True,False}\pValue{v} \colon \Frm[L_0] \to \{\True, \False \}source 23 inductively by: v¯()=False;v¯(pn)=v(pn);v¯(¬A)={Trueif v¯(A)=False;Falseotherwise.v¯(AB)={Trueif v¯(A)=True and v¯(B)=True;Falseif v¯(A)=False or v¯(B)=False.v¯(AB)={Trueif v¯(A)=True or v¯(B)=True;Falseif v¯(A)=False and v¯(B)=False.v¯(AB)={Trueif v¯(A)=False or v¯(B)=True;Falseif v¯(A)=True and v¯(B)=False.\pValue{v}(\lfalse) & = \False; \\ \pValue{v}(\Obj p_n) & = \pAssign{v}(\Obj p_n); \\ \pValue{v}(\lnot !A) & = \begin{cases} \True & \text{if } \pValue{v}(!A) = \False;\\ \False & \text{otherwise.} \end{cases} \\ \pValue{v}(!A \land !B) & = \begin{cases} \True & \text{if $\pValue{v}(!A) = \True$ and $\pValue{v}(!B) = \True$;}\\ \False & \text{if $\pValue{v}(!A) = \False$ or $\pValue{v}(!B) = \False$}. \end{cases}\\ \pValue{v}(!A \lor !B) & = \begin{cases} \True & \text{if $\pValue{v}(!A) = \True$ or $\pValue{v}(!B) = \True$;}\\ \False & \text{if $\pValue{v}(!A) = \False$ and $\pValue{v}(!B) = \False$}. \end{cases}\\ \pValue{v}(!A \lif !B) & = \begin{cases} \True & \text{if $\pValue{v}(!A) = \False$ or $\pValue{v}(!B) = \True$;}\\ \False & \text{if $\pValue{v}(!A) = \True$ and $\pValue{v}(!B) = \False$}. \end{cases}\\source 24

            source 21

            Exercise: A ternary connective

            Consider adding to L0\Lang{L_0}source 120 a ternary connective \diamondsuitsource 120 with evaluation given by v¯((A,B,C))={v¯(B)if v¯(A)=True;v¯(C)if v¯(A)=False.\pValue{v}(\diamondsuit( !A , !B , !C ) ) = \begin{cases} \pValue{v}( !B ) & \text{if $\pValue{v}(!A) = \True$;}\\ \pValue{v}( !C ) & \text{if $\pValue{v}(!A) = \False $}. \end{cases}source 122 Write down the truth table for this connective.

            source 119

            Theorem: Local determination

            Suppose that v1\pAssign{v_1}source 134 and v2\pAssign{v_2}source 135 are valuations that agree on the propositional variables occurring in A!Asource 136, i.e., v1(pn)=v2(pn)\pAssign{v_1}(\Obj p_n) = \pAssign{v_2}(\Obj p_n)source 136 whenever pn\Obj p_nsource 137 occurs in some formula A!Asource 138. Then v1¯\pValue{v_1}source 138 and v2¯\pValue{v_2}source 138 also agree on A!Asource 139, i.e., v1¯(A)=v2¯(A)\pValue{v_1}(!A) = \pValue{v_2}(!A)source 139.

            source 133

            Proof

            By induction on A!Asource 143.

            End of proof.

            Definition: Satisfaction by a valuation

            We can inductively define the notion of satisfaction of a formula A!Asource 148 by a valuation v\pAssign{v}source 149, vA\pSat{v}{!A}source 149, as follows. (We write vA\pSat/{v}{!A}source 150 to mean “not vA\pSat{v}{!A}source 150.”)

            Case: v\pSat/{v}{\indfrm}source 153.

            1. Case: vpi\pSat{v}{\indfrm}source 158 iff v(pi)=True\pAssign{v}(\Obj p_i) = \Truesource 159.

              Case: v¬B\pSat{v}{\indfrm}source 162 iff vB\pSat/{v}{!B}source 163.

              Case: vBC\pSat{v}{\indfrm}source 166 iff vB\pSat{v}{!B}source 166 and vC\pSat{v}{!C}source 167.

              Case: vBC\pSat{v}{\indfrm}source 170 iff vB\pSat{v}{!B}source 171 or vC\pSat{v}{!C}source 171 (or both).

              Case: vBC\pSat{v}{\indfrm}source 174 iff vB\pSat/{v}{!B}source 174 or vC\pSat{v}{!C}source 175 (or both).

            If Γ\Gammasource 182 is a set of formulas, vΓ\pSat{v}{\Gamma}source 182 iff vA\pSat{v}{!A}source 183 for every AΓ!A \in \Gammasource 183.

            source 146

            Proposition: Satisfaction agrees with truth value

            vA\pSat{v}{!A}source 187 iff v¯(A)=True\pValue{v}(!A) = \Truesource 187.

            source 186

            Proof

            By induction on A!Asource 191.

            End of proof.

            Exercise: Prove satisfaction agrees with truth value

            Prove Proposition: Satisfaction agrees with truth value

            source 194

            Semantic Notions

            We define the following semantic notions:

            Definition: Satisfiability, tautology, and semantic consequence

            1. A formula A!Asource 17 is satisfiable if for some v\pAssign{v}source 18, vA\pSat{v}{!A}source 18; it is unsatisfiable if for no v\pAssign{v}source 19, vA\pSat{v}{!A}source 19;

            2. A formula A!Asource 20 is a tautology if vA\pSat{v}{!A}source 20 for all valuations v\pAssign{v}source 21;

            3. A formula A!Asource 22 is contingent if it is satisfiable but not a tautology;

            4. If Γ\Gammasource 24 is a set of formulas, ΓA\Gamma \Entails !Asource 24 (“Γ\Gammasource 24 entails A!Asource 25”) if and only if vA\pSat{v}{!A}source 25 for every valuation v\pAssign{v}source 26 for which vΓ\pSat{v}{\Gamma}source 26.

            5. If Γ\Gammasource 27 is a set of formulas, Γ\Gammasource 27 is satisfiable if there is a valuation v\pAssign{v}source 28 for which vΓ\pSat{v}{\Gamma}source 29, and Γ\Gammasource 29 is unsatisfiable otherwise.

            source 15

            Exercise: Classify formulas semantically

            For each of the following four formulas determine whether it is (a) satisfiable, (b) tautology, and (c) contingent.

            1. (p0(¬p1¬p0))( \Obj p_0 \lif ( \lnot \Obj p_1 \lif \lnot \Obj p_0 ) )source 38.

            2. ((p0¬p1)(¬p0p2))((p2p0)(p0p1))( ( \Obj p_0 \land \lnot \Obj p_1 ) \lif ( \lnot \Obj p_0 \land \Obj p_2 )) \liff ( ( \Obj p_2 \lif \Obj p_0 ) \lif ( \Obj p_0 \lif \Obj p_1 ))source 39.

            3. (p0p1)(p2¬p1)( \Obj p_0 \liff \Obj p_1 ) \lif ( \Obj p_2 \liff \lnot \Obj p_1 )source 40.

            4. ((p0(¬p1p2))(p2(p0p1)))(( \Obj p_0 \liff ( \lnot \Obj p_1 \land \Obj p_2 )) \lor ( \Obj p_2 \lif ( \Obj p_0 \liff \Obj p_1 )))source 41.

            source 34

            Proposition: Basic semantic consequence facts

            1. A!Asource 48 is a tautology if and only if A\emptyset \Entails !Asource 49;

            2. If ΓA\Gamma \Entails !Asource 50 and ΓAB\Gamma \Entails !A \lif !Bsource 50 then ΓB\Gamma \Entails !Bsource 51;

            3. If Γ\Gammasource 52 is satisfiable then every finite subset of Γ\Gammasource 52 is also satisfiable;

            4. Monotonicity: if ΓΔ\Gamma \subseteq \Deltasource 54 and ΓA\Gamma \Entails !Asource 55 then also ΔA\Delta \Entails !Asource 55;

            5. Transitivity: if ΓA\Gamma \Entails !Asource 56 and Δ{A}B\Delta \cup \{ !A\} \Entails !Bsource 57 then ΓΔB\Gamma \cup \Delta \Entails !Bsource 57.

            source 45

            Proof

            Exercise.

            End of proof.

            Exercise: Prove the semantic consequence facts

            Prove Proposition: Basic semantic consequence facts

            source 66

            Proposition: Consequence and unsatisfiability

            ΓA\Gamma \Entails !Asource 71 if and only if Γ{¬A}\Gamma \cup \{\lnot !A\}source 71 is unsatisfiable.

            source 70

            Proof

            Exercise.

            End of proof.

            Exercise: Prove the consequence-unsatisfiability equivalence

            Prove Proposition: Consequence and unsatisfiability

            source 79

            Theorem: Semantic deduction theorem

            ΓAB\Gamma \Entails !A \lif !Bsource 84 if and only if Γ{A}B\Gamma \cup \{!A\} \Entails !Bsource 85.

            source 83

            Proof

            Exercise.

            End of proof.

            Exercise: Prove the semantic deduction theorem

            Prove Theorem: Semantic deduction theorem

            source 92