Summary
This lesson opens the fourth and last content unit of the course, devoted to metatheory. It presents its three lessons and the thread that joins them: the two previous units built, on the same language, two notions of what follows from some premises, deduction and semantic consequence , and this unit proves that they coincide, first for finite premises (soundness, Kalmár's lemma and completeness), then for arbitrary premises (the compactness theorem and strong completeness), and finally offers a third way of deciding consequence, the resolution method, of which it also proves that it is sound and complete. It explains why compactness is proved before strong completeness, and not the other way round; it shows with two examples the questions that neither the search for deductions nor the truth table can answer on its own; it specifies what is taken from the three previous units and gives advice for studying it.
Learning objectives
- Describe the sequence of the three lessons of the unit, explain which results of each one the next one uses and justify that compactness is proved, by a semantic argument, before strong completeness.
- Distinguish the two faces of the course, derivability , existential and finite, and semantic consequence , universal over infinitely many valuations, and state soundness and completeness in their three degrees.
- Explain why the coincidence of the two relations cannot be established case by case, but only with metatheorems, and recognize which instruments of the previous units (inductions, proof by cases, deduction theorems, coincidence lemma, conjunctive forms) each proof will use.
- Read deductions about deductions and semantic arguments of the unit, distinguishing at each step the relation being spoken of, and organize study starting from the entry assessment.
Entry assessment
Before working through the unit, it is worth measuring the starting point. This assessment is a self-assessment: it is not graded, and its only purpose is to guide your study. Its fifteen questions are of two kinds. The first six concern what the unit takes for granted from the previous ones: the definition of semantic consequence and its reduction to unsatisfiability, the value of joint denial, the coincidence lemma, the finiteness of deduction and proof by cases; anyone who misses one of them should review, before going on, the lesson that the question names. The remaining nine cover, in order, the three lessons of this unit, three each. Anyone who has never studied the metatheory of a logical system should not expect to get them right, and that does not matter: each answer, right or wrong, explains the point and names the lesson that deals with it. The same skills are measured again at the end, in the synthesis lesson.
Before starting the unit, measure your starting point. It is a short self-assessment: it is not graded and its result is stored only in your browser.
Question 1From earlier unitsWhat does mean?
That is , the syntactic relation of the unit on deduction. That the two relations coincide is precisely what this unit proves; taking them to be the same from the start is assuming what is to be proved.That is the satisfiability of . With and there is a valuation that satisfies both, and yet is not a consequence of : the valuation that gives to and to prevents it.Consequence may need several premises together: , and neither of the two premises, alone, has as a consequence.It is the definition of semantic consequence: a universal assertion about valuations, which does not mention deductions. If is empty, it says that is valid; if is unsatisfiable, it holds for every .Question 2From earlier unitsWhich of the following assertions is equivalent to ?
It is not enough: is satisfiable and is not a consequence of . Consequence reduces to the impossibility of falsifying the conclusion, not to the possibility of asserting it.It is exactly the negation of consequence: a valuation that satisfies that set is a countermodel.That says that is valid, which is the case in which is empty; with premises, a conclusion that is not valid may nevertheless be a consequence.A valuation that satisfied would satisfy and falsify , and that is exactly what consequence excludes. It is the consequence and unsatisfiability theorem, which compactness and resolution use at every step.Question 3From earlier unitsIf a valuation gives the value , what value does it give ?
That is how alternative denial, “not both”, , would behave. Joint denial, “neither… nor”, has the value only when both components have the value .It would be the value if joint denial looked only at its second component; the clause multiplies the two complements, and that of is .It confuses joint denial with disjunction, which has the value when some component has it; joint denial is precisely the negation of disjunction.By the clause of joint denial, , and the first factor is . It is the first of the three cases of Kalmár's lemma, which the first lesson translates into a deduction.Review: Semantics of propositional logic
Question 4From earlier unitsThe variables of are and . Two valuations and give the same values to and to , and opposite values to all the other variables. If , what can be asserted about ?
Differences on variables that do not occur in do not affect it; that is why a formula with variables has a table of rows, even though valuations are infinitely many.Inverting the value of variables absent from changes nothing; only inverting that of or that of could change its value.Agreement on the variables of suffices; those of lower index that do not occur in it are as irrelevant as those of higher index.It is the coincidence lemma: the value of a formula depends only on the values of its variables, which are finitely many. Compactness uses it in its last step, and Kalmár's lemma, so that one row of the table suffices.Review: Semantics of propositional logic
Question 5From earlier unitsLet be an infinite set of formulas such that . What can be asserted?
The premises cannot be suppressed: , and is not a theorem.A deduction has finitely many lines and cannot use infinitely many premises; besides, nothing obliges one to use all the premises available.Finiteness was proved in the unit on deduction by looking only at the length of deductions, without any semantic theorem. It is semantic consequence, not deduction, that needs compactness.It is the finiteness of deduction: a deduction is a finite sequence and uses only finitely many premises, which form that subset. Its semantic counterpart, that whatever is a consequence of is a consequence of a finite part of it, is not trivial: it is the compactness theorem.Review: Formal deductive systems
Question 6From earlier unitsIt is known that and that . What does proof by cases allow one to conclude?
There is no need to know it: proof by cases uses that one of the two alternatives holds, without deciding which. That deduces neither of the two is the usual situation, and the rule holds all the same.It does not follow: with empty and the two hypotheses hold, and is false when is false, so it should not be derivable.That is true, by monotonicity, but it is much less than what is known: proof by cases makes it possible to dispense with the assumption.Proof by cases eliminates the assumption: if is derivable both with and with , it is derivable without either of them. Weak completeness applies it once for each variable to eliminate the premises that Kalmár's lemma fixes.Question 7The course system is sound. What does that property assert?
That is the converse direction, completeness, which requires a much finer instrument, Kalmár's lemma.That is the consistency of the system, which follows from soundness but is weaker: the course system with the additional axiom is consistent, and it derives something that is not valid.It mixes two errors: it reverses the direction, and it confuses what is true under some valuation with what is valid. The variable is true under many valuations and is not a theorem.What is derivable from some premises is true in every valuation that makes them true. It is proved by strong induction on the number of lines, with one case for each axiom and for each rule.Review: Soundness and completeness
Question 8Let be the valuation that gives to and to . What does Kalmár's lemma assert, with the variables and , for the formula ?
It forgets the sign: the formula has the value in that row, and a deduction from premises that the row makes true cannot end in a formula that the row makes false.It takes the premise for wrongly: since , the row contributes , not ; with and true, the biconditional has the value .It suppresses the premises of the row: is not valid. The lemma derives from the row, and only when the formula has the value in all rows can the premises be eliminated.The row of becomes the premises and ; since has the value in it, the signed formula is its negation. The lemma asserts that the system derives from the row what the row says about the formula.Review: Soundness and completeness
Question 9Can a deductive system be sound without being complete?
It is a frequent error, and the first lesson of the unit refutes it. A system can derive only what it should without deriving everything it should, and the other way round.That the axioms are valid ensures that nothing false is derived, not that everything true is derived: a system without axioms is sound and derives no theorem.An inconsistent system derives a formula and its negation, and cannot be sound, because no valuation satisfies both.Its axioms are valid and modus ponens preserves truth; but the marking of the lesson on deductive systems showed that this tautology is not derivable in it. Soundness and completeness are independent properties.Review: Soundness and completeness
Question 10What does the compactness theorem assert?
It already fails with two formulas: and are satisfiable, and is not. What compactness examines are the finite subsets, not the formulas taken singly.The set is infinite and unsatisfiable.Every set has a satisfiable finite subset, the empty one; the condition must hold for all finite subsets, not for some.One direction is immediate; the other requires building a single valuation from those that satisfy, each on its own, the finite parts. The lesson does so by fixing one by one the values of .Review: The compactness theorem
Question 11Let be an infinite set of formulas such that . What does compactness ensure?
A finite part may need several formulas: from follows , and no formula of that set, alone, has it as a consequence.The empty subset is finite, and only tautologies follow from it; compactness ensures some finite subset, not all.The premises may be indispensable: , and is not valid.It is the consequence form: whatever follows from an infinite set already follows from a finite part of it. For deduction, it was finiteness, a triviality; for consequence, it is a theorem.Review: The compactness theorem
Question 12The first lesson proves completeness for finite sets of premises. Why does the finiteness of deduction not suffice to extend it to an infinite set ?
The argument is circular: finiteness applies to deductions, and from it is not yet known that any exists.No formula has infinitely many variables. The obstacle lies not in the formulas, but in the fact that infinitely many premises cannot be discharged into a single implication.It is true: the second lesson proves it, with compactness. What falls short is the method of the first lesson, not the statement.To pass from to one must first find a finite with ; finiteness would give a finite part only if it were already known that , which is what is sought.Review: The compactness theorem
Question 13What is the resolvent of the clauses and ?
It joins the clauses without removing the complementary pair: that is not a resolvent, but a tautological clause, which says nothing.It loses the literal of the second clause: the resolvent keeps everything that is not the eliminated pair, from both premises.They must not share a literal, but contain one and its opposite: in the first and in the second.The clauses clash on and : that pair is removed and the remaining literals are joined. It is proof by cases on the value of , written on sets.Review: The resolution method
Question 14To decide by resolution whether , which set of clauses does one try to refute?
It adds the conclusion instead of its negation; that set is satisfiable, and its satisfiability says nothing about consequence.Without the negation of the conclusion, the set of premises is satisfiable, and resolution has nothing to refute.It negates the premises, which must be kept: only the conclusion is negated.Consequence is equivalent to the unsatisfiability of the premises together with the negation of the conclusion; one writes its clausal form and looks for a refutation, which here has two steps.Review: The resolution method
Question 15Can the empty clause be derived by resolution from the set ?
The rule eliminates a single pair per step, and not by convention: eliminating two would derive from a satisfiable set.The first step gives , and no sequence of steps reaches , because the set is satisfiable and everything derived is a consequence of it.The reason is another: the set of the four clauses of two literals in and has no unit clauses and is unsatisfiable, and resolution refutes it by deriving first and .The valuation that gives to and to satisfies it; if were derived, it would be unsatisfiable. The two legitimate resolvents, and , are tautological.Review: The resolution method
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