Part I
Propositional Logic
1
Chapter 1
Propositional Logic
1.1
prp:prp:pre:
sec
Preliminaries
The language L0 of classical propositional logic comprises as basic symbols
(1) countably many propositional variables p0 , p1 , . . . , as well as (2) symbols
for the logical logical operators ¬ (negation), ∧ (conjunction), ∨ (disjunction),
→ (conditional), ↔ (biconditional), the propositional constant for falsity ⊥,
the propositional constant for truth >, and the two parentheses ( and ). We
assume that these symbols are all distinct and no one occurs as a part of
another one. We refer to the set of the propositional variables as At0 .
Definition 1.1 (Formulas). The set Frm(L0 ) of the formulas of the language
L0 is inductively defined as the smallest set of strings over the alphabet containing At0 and such that if ϕ and ψ are in Frm(L0 ), then so are:
1. (¬ϕ);
2. (ϕ → ψ).
prp:prp:pre:
thm:induction
Theorem 1.2. Principle of induction on formulas: If some property P holds
of all the propositional variables and is such that it holds for (¬ϕ) and (ϕ → ψ)
whenever it holds for ϕ and ψ, then P holds of all formulas in Frm(L0 ) .
Proof. Let S be the collection of all formulas with property P , so that, in
particular, S ⊆ Frm(L0 ). Then S contains the propositional variables and is
closed under the logical operators; since Frm(L0 ) is the smallest such class,
also Frm(L0 ) ⊆ S. So Frm(L0 ) = S, and every formula has propery P .
Problem 1.1. Prove that any formula in Frm(L0 ) is balanced, in that it has
as many left parentheses as right ones.
Problem 1.2. Prove that no formula begins with ¬ and that no proper initial
segment of a formula is a formula.
3
explanation
The formulas (ϕ ∨ ψ) and (ϕ ∧ ψ) abbreviate ((¬ϕ) → ψ) and ¬(ϕ → (¬ψ)),
respectively. Similarly, ϕ ≡ ψ abbreviates (ϕ → ψ) ∧ (ψ → ϕ). Parentheses
around ¬ϕ are usually dropped, with the understanding that ¬ binds the shortest formula that follows it; outermost parentheses are likewise usually dropped.
Proposition 1.3 (Unique Readability). Any formula ϕ in Frm(L0 ) has exactly
one parsing as one of the following
1. pn for some pn ∈ At0
2. (¬ψ) for some ψ in Frm(L0 );
3. (ψ → θ) for some ψ, θ in Frm(L0 ).
Moreover, such parsing is unique, in that, e.g., ϕ cannot have the form (¬ψ)
in two different ways.
Proof. By induction on ϕ. For instance, suppose that ϕ has two distinct readings as (ψ → θ) and (ψ 0 → θ0 ). Then ψ and ψ 0 must be the same (or else one
would be a proper initial segment of the other); so is the two readings of ϕ are
distinct it must be because θ and θ0 are distinct readings of the same sequence
of symbols, which is impossible by the inductive hypothesis.
Theorem 1.4 (Principle of Definition by Recursion). For any set V and
function symbols i : At0 → V and h1 , h2 from V and V × V , respectively, into
V , there exists exactly one function f : Frm(L0 ) → V satisfying the following
equations:
f (pn ) = i(pn )
f (¬ϕ) = h1 (f (ϕ))
f (ϕ → ψ) = h2 (f (ϕ), f (ψ))
Proof. Let F be the class of all functions g such that:
• dom(g) is finite and closed under subformulas;
• whenever ϕ is in dom(g) then g satisfies the equation corresponding to
ϕ.
S
Put f = F. It is easy to see that:
1. No two functions g and g 0 in F disagree on any of the arguments on
which they are both defined. Hence, f is well-defined as a function.
(This requires unique readability.)
2. f satisfies the equations.
3. f is unique.
4. f is total, i.e., dom(f ) = Frm(L0 ).
4
propositional-logic by OLP / CC–BY
prp:prp:pre:
thm:rec
These are established by induction on formulas.
Definition 1.5 (Uniform Substitution). If ϕ and ψ are formulas, and pi is a
propositional variable, then ϕ[ψ/pi ] denotes the result of replacing each occurrence of pi by an occurrence of ψ in ϕ; similarly, the simultaneous substitution
of p1 , . . . , pn by formulas ψ1 , . . . , Bn is denoted by ϕ[ψ1 /p1 ], . . . , [ψn /pn ].
Problem 1.3. Give a mathematically rigorous definition of ϕ[ψ/pi ] using Theorem 1.4.
1.2
Semantics for Propositional Logic
prp:prp:sem:
sec
Definition 1.6 (Valuations). Let {1, 0} be the set of the two truth values,
“true” and “false.” A valuation for L0 is a function v assigning either 1 or 0
to the propositional variables of the language, i.e., v : At0 → {1, 0}.
Theorem 1.7. Every valuation v can be “lifted” to a unique function v : Frm(L0 ) →
{1, 0} such that:
v(pn ) = v(pn );
(
1 if v(ϕ) = 0;
v(¬ϕ) =
0 otherwise.
(
1 if v(ϕ) = 0 or v(ψ) = 1;
v(ϕ → ψ) =
0 if v(ϕ) = 1 and v(ψ) = 0.
Proof. Apply Theorem 1.4.
Problem 1.4. Show that the valuation clauses for → and ¬ give the right
truth tables for ∧ and ∨:
ψ
1
1
0
0
prp:prp:sem:
thm:LocalDetermination
ψ
1
0
1
0
ϕ∧ψ
1
0
0
0
ϕi ∨ ψ
1
1
1
0
Theorem 1.8 (Local Determination). Suppose that v1 and v2 are valuations
that agree on the propositional letters occurring in ϕ, i.e., v1 (pn ) = v2 (pn )
whenever pn occurs in ϕ. Then they also agree on any ϕ, i.e., v 1 (ϕ) = v 2 (ϕ).
Proof. By induction on ϕ.
Definition 1.9. Define the following semantic notions:
1. A formula ϕ is satisfiable if for some v, v(ϕ) = 1; it is unsatisfiable if for
no v, v(ϕ) = 1;
propositional-logic by OLP / CC–BY
5
2. A formula ϕ is a tautology if v(ϕ) = 1 for all valuations v;
3. A formula ϕ is contingent if it is satisfiable but not a tautology;
4. if Γ is a set of formulas, Γ ϕ (“Γ entails ϕ”) if and only if v(ϕ) = 1 for
every valuation v on which v(ψ) = 1 for every ψ ∈ Γ .
5. if Γ is a set of formulas, Γ is satisfiable if there is a valuation v on which
v(ψ) = 1 for every ψ ∈ Γ , and Γ is unsatisfiable otherwise.
Proposition 1.10. The following can all be proved with little more than “definition chasing”:
prp:prp:sem:
prop:semanticalfacts
1. ϕ is a tautology if and only if ∅ ϕ;
2. Γ ϕ if and only if Γ ∪ {¬A} is unsatisfiable;
3. if Γ ϕ and Γ ϕ → ψ then Γ ψ;
4. if Γ is satisfiable then every finite subset of Γ is also satisfiable;
5. Monotony: if Γ ⊆ ∆ and Γ ϕ then also ∆ ϕ;
6. Cut: if Γ ϕ and ∆ ∪ {ϕ} ψ then Γ ∪ ∆ ψ;
7. Semantic deduction theorem: Γ ϕ ⊇ ψ if and only if Γ ∪ {ϕ} ψ.
prp:prp:sem:
def:Monotony
prp:prp:sem:
def:Cut
prp:prp:sem:
def:DeductionTheorem
Proof. Exercise.
Problem 1.5. Prove Proposition 1.10
1.3
Axiomatic Deduction
prp:prp:axd:
sec
Definition 1.11 (Axioms). The set of Ax0 of axioms of propositional logic
comprises all formulas of the following forms:
[Ax1] ϕ → (ψ → ϕ);
[Ax2] (ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ));
[Ax3] (¬ϕ → ¬ψ) → (ψ → ϕ).
Definition 1.12 (Derivability). If Γ is a set of formulas of L0 and ϕ a formula,
then a derivation of ϕ from Γ is a finite sequence ϕ1 , . . . , ϕn of formulas such
that ϕn = ϕ and for each i ≤ n one of the following holds:
1. ϕi ∈ Γ ; or
2. ϕi is an axiom; or
3. there are j, k < i so that ϕj is ϕk → ϕi .
6
propositional-logic by OLP / CC–BY
The last clause just says that ϕi can be obtained from previously occurring
ϕk and ϕk → ϕi by modus ponens (MP). We write Γ ` ϕ (“Γ derives ϕ,” or
“ϕ is derivable from Γ ”) to mean that there is a derivation of ϕ from Γ . When
Γ is empty, we write ` ϕ to mean ∅ ` ϕ.
prp:prp:axd:
prop:easyproofs
prp:prp:axd:
prop:easyproofs:trans
prp:prp:axd:
prop:easyproofs:refl
Proposition 1.13. The following are derivable:
1. Transitivity: {ϕ → ψ, ψ → χ} ` ϕ → χ.
2. ` ϕ → ϕ.
Proof. For part (1), the following instances of Ax1 and Ax2 are needed:
(ψ → χ) → (ϕ → (ψ → χ));
(ϕ → (ψ → χ)) → ((ϕ → ψ) → (ϕ → χ)).
For part (2), consider the instance of Ax1: ϕ → ((ϕ → ϕ) → ϕ) and distribute
the outermost implication by Ax2.
prp:prp:axd:
prop:contain
Proposition 1.14. For any set Γ of formulas:
1. If ϕ is an axiom then Γ ` ϕ;
prp:prp:axd:
prop:contain:reflex
prp:prp:axd:
prop:contain:monotony
2. If ϕ is in Γ then Γ ` ϕ;
3. Monotony: if Γ ⊆ ∆ and Γ ` ϕ then also ∆ ` ϕ;
4. Γ ` ϕ if and only if there is a finite subset Γ0 of Γ such that Γ0 ` ϕ.
Theorem 1.15. Let Thm(Γ ) = {ϕ : Γ ` ϕ}. Then Thm(Γ ) is the smallest
set of formulas containing the axioms, every formula in Γ , and is closed under
modus ponens (from ϕ and ϕ → ψ infer ψ).
Proof. We know from Proposition 1.14 that Thm(Γ ) has the desired properties;
so we need to show that it is the smallest such. Let ∆ be any other set of
formulas containing the axioms and Γ and closed under modus ponens. Prove
that Thm(Γ ) ⊆ ∆ by induction on the length of a proof of ϕ from Γ .
prp:prp:axd:
cor:induction-thms
prp:prp:axd:
thm:deduction-thm
Corollary 1.16 (Principle of induction on theorems). Any property P that
holds of the axioms, of formulas in Γ , and is preserved by modus ponens holds
of every formula in Thm(Γ ).
Theorem 1.17 (Deduction Theorem). Γ ∪ {ϕ} ` ψ if and only if Γ ` ϕ → ψ.
Proof. The “if” direction is immediate; if Γ ` ϕ → ψ then also Γ ∪ {ϕ} ` ϕ →
ψ, so there is a derivation of ϕ → ψ from Γ ∪ {ϕ}, and one more application
of modus ponens gives Γ ∪ {ϕ} ` ψ.
For the converse, proceed by induction on theorems. If ψ ∈ Γ or ψ is
an axiom then also Γ ` ψ → (ϕ → ψ) by Ax1, and modus ponens gives
Γ ∪ ` ϕ → ψ; and if ψ ∈ {ϕ} then Γ ` ϕ → ψ because the last sentence is the
same as ϕ → ϕ.
propositional-logic by OLP / CC–BY
7
explanation
For the inductive step, suppose ψ is obtained by modus ponens from χ → ψ
and χ. Then Γ ∪ {ϕ} ` χ → ψ and Γ ∪ {ϕ} ` χ. By inductive hypothesis,
both
Γ ` ϕ → (χ → ψ);
Γ ` ϕ → χ.
But also
Γ ` (ϕ → (χ → ψ)) → ((ϕ → χ) → (ϕ → ψ)),
by Ax2, and two applications of modus ponens give Γ ` ϕ → ψ, as required.
Notice how Ax1 and Ax2 were chosen precisely so that the Deduction
Theorem would hold. The following proposition collects useful facts about
derivability that will be needed in the next section.
Proposition 1.18. The following are all derivable:
1. ` (ϕ → ψ) → ((ψ → χ) → (ϕ → χ);
prp:prp:axd:
derivfacts:a
2. Contraposition: if Γ ∪ {¬ϕ} ` ¬ψ then Γ ∪ {ψ} ` ϕ;
3. Ex Falso Quodlibet: {ϕ, ¬ϕ} ` ψ;
prp:prp:axd:
derivfacts:d
5. if Γ ` ¬¬ϕ then Γ ` ϕ;
prp:prp:axd:
derivfacts:e
Proof. Part (1) follows from Proposition 1.13, part (1) by two applications of
the Deduction Theorem. For part (2):
1. Γ ∪ {¬ϕ} ` ¬ψ
hyp.
2. Γ ∪ ` ¬ϕ → ¬ψ
Ded. Thm.
3. Γ ` (¬ϕ → ¬ψ) → (ψ → ϕ) Ax3 and monotony
4. Γ ` ψ → ϕ
MP
5. Γ ∪ {ψ} ` ϕ
Ded. Thm.
For part (3) we have {¬ϕ, ¬ψ} ` ¬ϕ by Prop Proposition 1.14, part (2),
and {ϕ, ¬ϕ} ` ψ follows by (2). Part (4): since {¬¬ϕ, ¬ϕ} ` ¬¬¬ϕ by ex falso
quodlibet, apply (2). Now for part (5), Γ ` ¬¬ϕ ⊃ ϕ by the previous part, the
deduction theorem, and monotony, so if Γ ` ¬¬ϕ also Γ ` ϕ by MP.
Theorem 1.19 (Cut). If Γ ` ϕ and ∆, ϕ ` χ then Γ ∪ ∆ ` χ.
8
prp:prp:axd:
derivfacts:b
prp:prp:axd:
derivfacts:c
4. Double Negation: {¬¬ϕ} ` ϕ;
Proof.
1.
2.
3.
4.
5.
6.
prp:prp:axd:
prop:derivfacts
The following shows that the conclusion is derivable:
Γ `ϕ
hyp.
Γ ∪∆`ϕ
monotony, 1.
∆, ϕ ` χ
hyp.
Γ ∪ ∆, ϕ ` χ
monotony, 3.
Γ ∪ ∆ ` ϕ → χ Ded. Thm., 4.
Γ ∪∆`χ
MP 2, 5.
propositional-logic by OLP / CC–BY
“Monotony” refers to Proposition 1.14(3).
prp:prp:axd:
lem:trivial
Lemma 1.20. ϕ → ψ, ϕ ` ψ.
Proof. Clearly ϕ → ψ ` ϕ → ψ. Apply the Deduction Theorem.
prp:prp:axd:
lem:notphi
Lemma 1.21. ϕ → ¬ϕ ` ¬ϕ.
Proof. The following shows the conclusion is derivable.
1.
2.
3.
4.
5.
6.
7.
8.
9.
prp:prp:axd:
prop:consistent
prp:prp:axd:
prop:consistent:0
prp:prp:axd:
prop:consistent:1
prp:prp:axd:
prop:consistent:2
prp:prp:axd:
prop:consistent:3
¬¬ϕ ` ϕ
¬¬¬¬ϕ ` ¬¬ϕ
¬¬(ϕ → ¬ϕ) ` ϕ → ¬ϕ
¬ϕ ` ¬¬¬ϕ
¬¬(ϕ → ¬ϕ), ϕ ` ¬ϕ
¬¬ϕ, ¬¬(ϕ → ¬ϕ) ` ¬ϕ
¬¬ϕ, ¬¬(ϕ → ¬ϕ) ` ¬¬¬ϕ
¬¬ϕ ` ¬(ϕ → ¬ϕ)
(ϕ → ¬ϕ) ` ¬ϕ
Double negation
Double negation
Double negation
contraposition, 2
Deduction theorem, 3
Cut, 1, 5
Cut 4, 6
Contraposition, 7
contraposition, 8.
Proposition 1.22. The following hold:
1. ψ ` ¬¬ψ
2. ϕ → ¬ψ, ψ ` ¬ϕ
3. ϕ → ψ, ϕ → ¬ψ ` ¬ϕ.
4. If Γ ∪ {ϕ} ` ψ and Γ ∪ {ϕ} ` ¬ψ then Γ ` ¬ϕ.
Proof. For part (1), we have ¬¬¬ψ ` ¬ψ by double negation, whence ψ ` ¬¬ψ
by Proposition 1.18, part (2). For part (2) we have:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
` (¬¬ϕ → ¬¬¬ψ) → (¬¬ψ → ¬ϕ)
¬¬ϕ → ¬¬¬ψ, ¬¬ψ ` ¬ϕ
ψ ` ¬¬ψ
¬¬ϕ → ¬¬¬ψ, ψ ` ¬ϕ
¬ψ → ¬¬¬ψ, ϕ → ¬ψ ` ϕ → ¬¬¬ψ
` ¬ψ → ¬¬¬ψ
ϕ → ¬ψ ` ϕ → ¬¬¬ψ
` ¬¬ϕ → ϕ
¬¬ϕ → ϕ, ϕ → ¬¬¬ψ ` ¬¬ϕ → ¬¬¬ψ
ϕ → ¬¬¬ψ ` ¬¬ϕ → ¬¬¬ψ
ϕ → ¬ψ ` ¬¬ϕ → ¬¬¬ψ
ϕ → ¬ψ, ψ ` ¬ϕ
Ax3
1, Ded. Thm., twice
Double Neg.
2, 3, Cut
Transitivity (Proposition 1.13, part (1))
Double Neg. + Ded. Thm.
5, 6, Cut
Double. Neg + Ded. Thm.
Transitivity
8, 9, Cut
7, 10, Cut
4, 11, Cut.
Part (3) follows because:
propositional-logic by OLP / CC–BY
9
1.
2.
3.
4.
5.
6.
ϕ → ψ, ϕ ` ψ
ϕ → ¬ψ, ψ ` ¬ϕ
ϕ → ψ, ϕ → ¬ψ, ϕ ` ¬ϕ
ϕ → ψ, ϕ → ¬ψ ` ϕ → ¬ϕ
ϕ → ¬ϕ ` ¬ϕ
ϕ → ψ, ϕ → ¬ψ ` ¬ϕ
Lemma 1.20
part (2)
1, 2, Cut.
deduction theorem
lemma Lemma 1.21
4, 5, Cut.
Finally, for part (4): from the hypotheses by the deduction theorem, Γ `
ϕ → ψ and Γ ` ϕ → ¬ψ; from part (3) and monotony, Γ, ϕ → ψ, ϕ → ¬ψ `
¬ϕ; two applications of Cut give the desired result.
Definition 1.23 (Consistency). A set Γ of formulas is consistent if and only
if Γ 0 ⊥; it is inconsistent otherwise.
Proposition 1.24. Γ is inconsistent iff Γ ` ϕ for every sentence ϕ.
prp:prp:axd:
prop:incons
Proof. Exercise.
Problem 1.6. Prove Proposition 1.24
Proposition 1.25. Γ ` ϕ iff Γ ∪ {¬ϕ} is inconsistent.
prp:prp:axd:
prop:prov-incons
Proof. If Γ ` ϕ then also Γ, ¬ϕ ` ϕ by monotony, and Γ, ¬ϕ ` ¬ϕ by reflexivity, so Γ ∪ {¬ϕ} is inconsistent. Conversely, suppose Γ ∪ {¬ϕ} is inconsistent.
Then Γ ¬ϕ ` χ and Γ ¬ϕ ` ¬χ for some χ. By Proposition 1.22(4), Γ ` ¬¬ϕ.
But also ¬¬ϕ ` ϕ, by double negation so that by cut, Γ ` ϕ.
Proposition 1.26. If Γ is consistent, then for any formula ϕ, either Γ ∪ {ϕ}
is consistent or Γ ∪ {¬ϕ} is consistent.
prp:prp:axd:
prop:phi
Proof. From Proposition 1.25 we have that if Γ ∪ {¬ϕ} is inconsistent, then
Γ ` ϕ; if Γ ∪ {ϕ} is also inconsistent, then Γ ∪ {ϕ} ` ψ for any ψ. But then
by Cut Γ ` ψ for any ψ, so Γ is inconsistent.
Proposition 1.27. If Γ ` ϕ iff for some finite Γ0 ⊆ Γ , Γ0 ` ϕ.
Proof. If Γ ` ϕ, then there is a finite sequence of formulas ϕ1 , . . . , ϕn so that
ϕ ≡ ϕn and each ϕi is either a logical axiom, an element of Γ or follows from
previous formulas by modus ponens. Take Γ0 to be those ϕi which are in Γ .
Then the derivation is likewise a derivation from Γ0 , and so Γ0 ` ϕ. The other
direction is obvious.
10
propositional-logic by OLP / CC–BY
prp:prp:axd:
prop:proves-compact
1.4
Soundness of Propositional Logic
prp:prp:snd:
sec
prp:prp:snd:
thm:soundness
Theorem 1.28 (Soundness). If Γ ` ϕ then Γ ϕ.
Proof. By induction on theorems. If ϕ is an axiom then ϕ and hence a fortiori
Γ ϕ. Similarly if ϕ ∈ Γ , Γ ϕ. For the inductive step, suppose ϕ is obtained
by modus ponens from χ and χ → ϕ; also assume as inductive hypothesis that
the theorem holds for χ → ϕ and χ. The third item in Proposition 1.10 gives
the desired result.
Corollary 1.29. If Γ is satisfiable, then Γ is consistent. Hence, propositional
logic is consistent.
1.5
Completeness of Propositional Logic
prp:prp:com:
sec
prp:prp:com:
def:MaxCon
prp:prp:com:
lem:truth
prp:prp:com:
lem:truth-1
prp:prp:com:
lem:truth-2
prp:prp:com:
lem:truth-3
Definition 1.30. A set Γ of formulas is maximally consistent if it is consistent
and if ∆ is a consistent set such that Γ ⊆ ∆ then Γ = ∆.
Lemma 1.31 (Truth Lemma). Let Γ be maximally consistent; then:
1. ϕ ∈ Γ if and only if ¬ϕ ∈
/ Γ;
2. Γ ` ϕ if and only if ϕ ∈ Γ ;
3. ϕ → ψ ∈ Γ if and only if either ϕ ∈
/ Γ or ψ ∈ Γ .
Proof.
1. Suppose ϕ ∈ Γ . If also ¬ϕ ∈ Γ , then Γ is inconsistent; and if
neither ϕ nor ¬ϕ is in Γ then by Proposition 1.26 one of Γ ∪ {ϕ} and
Γ ∪ {¬ϕ} is consistent, which means that Γ is not maximal. The other
direction is proved the same way.
2. If ϕ ∈ Γ then of course Γ ` ϕ. On the other hand, suppose Γ ` ϕ. If
ϕ∈
/ Γ then, by (1), ¬ϕ ∈ Γ . So Γ ` ϕ and Γ ` ¬ϕ, a contradiction.
3. Exercise.
The left-to-right direction of Lemma 1.31(2) shows that Γ is deductively
closed, i.e., if Γ ` ϕ then ϕ ∈ Γ for all ϕ.
Problem 1.7. Complete the proof of Lemma 1.31.
prp:prp:com:
thm:completeness
Theorem 1.32 (Completeness). If Γ is consistent then Γ is satisfiable.
propositional-logic by OLP / CC–BY
11
Proof. Let ϕ0 , ϕ1 , . . . be an exhaustive listing of all the formulas of the language. We recursively define an increasing sequence of sets of formulas Γ0 ,
Γ1 , . . . , by putting:
Γ0 = Γ
(
Γn ∪ {ϕn }
Γn+1 =
Γn ∪ {¬ϕn }
if Γn ∪ {ϕn } is consistent;
otherwise.
Then define:
Γ∗ =
[
Γn .
0≤n
The proof now proceeds by establishing, in turn, the following facts:
1. For each n, the set Γn is consistent (by induction on n, using Proposition 1.26);
2. Γ ∗ is consistent;
3. Γ ∗ is maximal.
Then define a valuation v by putting v(pi ) = 1 if and only if pi ∈ Γ ∗ . By
induction on ϕ it is then shown that membership in Γ ∗ coincides with truth
according to v also for more complex sentences: v(ϕ) = 1 if and only if ϕ ∈
Γ ∗ . In particular, v satisfies Γ ∗ , and since Γ ⊆ Γ ∗ , also Γ is satisfiable, as
desired.
Corollary 1.33. If Γ ϕ then Γ ` ϕ.
Proof. If Γ 0 ϕ then Γ ∪ {¬ϕ} is consistent, by Proposition 1.25. So by the
completeness theorem, Γ ∪ {¬ϕ} is satisfiable, and Γ 2 ϕ.
Proposition 1.34 (Compactness Theorem).
every finite subset Γ0 of Γ is satisfiable.
Γ is satisfiable if and only if
Proof. Γ is unsatisfiable if and only if it is inconsistent, if and only if some
finite subset Γ0 of Γ is inconsistent, if and only if some finite subset Γ0 of Γ is
unsatisfiable.
Corollary 1.35. Γ ϕ if and only if for some finite subset Γ0 of Γ , Γ0 ϕ.
12
propositional-logic by OLP / CC–BY
prp:prp:com:
prop:compactness