Sets and Logic (234293) WS 2014/5
Lecture 7
Lecture 7
What did we do so far ?
• We have defined WFF.
• We have discussed the unique readability of formulas (words) of WFF.
• We have defined a meaning function for WFF with values in {0, 1}.
• We have introduced truth tables for formulas.
• We have shown truth table completeness of WFF with our meaning
functions.
7s-2014
1
Sets and Logic (234293) WS 2014/5
Lecture 7
Outline of lecture 7
We shall introduce
• Satisfiability and validity of formulas.
• Logical equivalence of formulas.
• The notion of logical consequence of formulas.
• Substitution of formulas in formulas.
• Various normal forms of formulas.
• Some history of the Propositional Calculus
• Why attend lectures
Back to outline of Lecture 7
7s-2014
2
Sets and Logic (234293) WS 2014/5
Lecture 7
Generality of semantic concepts
Validity,
Logical Equivalence and
Logical Consequence
are fundamental concepts of the
semantics of formal languages
in the most general sense.
We now introduce them for Propositional Logic,
but the student should have in mind
that they are really concepts about meaning functions.
7s-2014
3
Sets and Logic (234293) WS 2014/5
Lecture 7
Definition 1 (3.2.1)
Basic Semantic Concepts: Validity and inconsistency
• We say that a formula φ ∈ WFF is
(logically) valid or a tautology
if for every propositional assignment z
MP L(φ, z) = 1.
• We say that a formula φ ∈ WFF is a
contradiction or inconsistent
if for every propositional assignment z
MP L(φ, z) = 0.
7s-2014
4
Sets and Logic (234293) WS 2014/5
Lecture 7
Definition 2 (3.2.1, contd)
Basic Semantic Concepts: Satisfiability
• We say that a formula φ ∈ WFF is
satisfiable if there is a propositional
assignment z such that MP L(φ, z) = 1.
• We say that a set of formulas Σ ⊆ WFF
is satisfiable if there is
a propositional assignment z
such that MP L(φ, z) = 1 for every φ ∈ Σ.
We abreviate this as MP L(Σ, z) = 1.
Strictly speaking, we extend the function MP L to the power set of WFF.
7s-2014
5
Sets and Logic (234293) WS 2014/5
Lecture 7
Examples 3 (3.2.2)
(i) φ is valid if and only if ¬φ is not satisfiable.
(ii) φ is a contradiction if and only if ¬φ is valid.
(iii) Find an infinite set of valid formulas.
(iv) Find an infinite set of formulas which are contradictions.
(v) Find an infinite set of satisfiable formulas which are not valid.
7s-2014
6
Sets and Logic (234293) WS 2014/5
Lecture 7
Proposition-Exercise 4 (3.2.3)
Three important valid formulas
The following formulas φ ∈ WFF are valid:
(ψ1 → (ψ2 → ψ1))
((ψ1 → (ψ2 → ψ3)) → ((ψ1 → ψ2) → (ψ1 → ψ3)))
(((ψ1 → F) → F) → ψ1)
Those formulas will be used later as axioms
in the definition of proof sequences.
Back to outline of Lecture 7
7s-2014
7
Sets and Logic (234293) WS 2014/5
Lecture 7
Definition 5 (3.2.4)
Basic Semantic Concepts: Logical Equivalence
We say that two formulas φ1, φ2 are
logically equivalent or semantically equivalent
if and only if for every
propositional assignement z
MP L(φ1, z) = MP L(φ2, z).
7s-2014
8
Sets and Logic (234293) WS 2014/5
Lecture 7
Examples 6
The following pairs of formulas are equivalent:
• ¬φ and (φ → F).
• (¬φ ∨ ψ) and (φ → ψ).
• (φ ∨ ψ) and ((φ → F) → ψ).
• (φ ∧ ψ) and ¬(¬φ ∨ ¬ψ).
Use this to show the functional completeness of {→, F}, i.e.:
Proposition-Exercise 7
For every truth table T T there as a formula φ ∈ WFF{→,F}
such that T T = T Tφ .
7s-2014
9
Sets and Logic (234293) WS 2014/5
Lecture 7
Examples 8 (3.2.5)
• φ ∈ WFF is valid if and only if
φ is logically equivalent to the formula T.
• φ ∈ WFF is a contradiction if and only if
φ is logically equivalent to the formula F.
• φ ∈ WFF is valid if and only if
¬φ is logically equivalent to the formula F.
7s-2014
10
Sets and Logic (234293) WS 2014/5
Lecture 7
Proposition-Exercise 9 (3.2.6)
(i) φ is a tautology if and only if
T Tφ is the constant function with value 1.
(ii) φ is satisfiable if and only if there are
x1 , x2 , ..., xn ∈ {0, 1} such that
T Tφ (x1 , x2 , ..., xn ) = 1.
(iii) Two well formed formulas φ, ψ ∈ WFF are
logically equivalent if and only if
they have the same truth tables
associated with them,
i.e. T Tφ = T Tψ .
7s-2014
11
Sets and Logic (234293) WS 2014/5
Lecture 7
Proposition-Exercise 10 (3.2.7)
Show that the following pairs of formulas φ1 , φ2 are logically equivalent:
Commutativity:
φ1 = (ψ1 ∧ ψ2 ),
φ2 = (ψ2 ∧ ψ1 );
φ1 = (ψ1 ∨ ψ2 ),
φ2 = (ψ2 ∨ ψ1 );
Associativity:
7s-2014
φ1 = ((ψ1 ∧ ψ2 ) ∧ ψ3 ),
φ2 = ((ψ1 ∧ (ψ2 ∧ ψ3 ));
φ1 = ((ψ1 ∨ ψ2 ) ∨ ψ3 ),
φ2 = ((ψ1 ∨ (ψ2 ∨ ψ3 ));
12
Sets and Logic (234293) WS 2014/5
Lecture 7
Proposition-Exercise 11 (3.2.7, contd)
Show that the following pairs of formulas φ1 , φ2 are logically equivalent:
Distributivity:
φ1 = ((ψ1 ∧ ψ2 ) ∨ ψ3 ),
φ2 = ((ψ1 ∨ ψ3 ) ∧ (ψ2 ∨ ψ3 ));
φ1 = ((ψ1 ∨ ψ2 ) ∧ ψ3 ),
φ2 = ((ψ1 ∧ ψ3 ) ∨ (ψ2 ∧ ψ3 ));
De Morgan’s Laws:
φ1 = ¬(ψ1 ∧ ψ2 ),
φ2 = (¬ψ1 ∨ ¬ψ2 );
φ1 = ¬(ψ1 ∨ ψ2 ),
φ2 = (¬ψ1 ∧ ¬ψ2 );
Double negation:
φ1 = ¬¬ψ,
φ2 = ψ.
Back to outline of Lecture 7
7s-2014
13
Sets and Logic (234293) WS 2014/5
Lecture 7
Definition 12 (3.2.8)
Basic Semantic Concepts: Logical Consequence
Let Σ be a (possibly infinite) set of well formed formulas in WFF,
and let φ ∈ WFF.
We say that φ is a logical (semantical) consequence of Σ
or alternatively
Σ logically (semantically) entails φ
if and only if for every propositional assignement z such that
MP L(Σ, z) = 1
we have also that
MP L(φ, z) = 1.
We write Σ |= φ for Σ entails φ.
7s-2014
14
Sets and Logic (234293) WS 2014/5
Lecture 7
Examples 13 (3.2.9)
(i) φ is valid if and only if
the empty set ∅ entails φ, i.e. ∅ |= φ;
(ii) {φ} |= φ;
(iii) more generally, if φ ∈ Σ, then Σ |= φ.
(iv) φ and ψ are logically equivalent,
if and only if
{φ} |= ψ and {ψ} |= φ.
7s-2014
15
Sets and Logic (234293) WS 2014/5
Lecture 7
Proposition-Exercise 14
Some simple but useful properties
of the logical consequence relation.
(i) (False implies everything)
For every φ ∈ WFF we have that {F} |= φ;
(ii) For every φ, ψ ∈ WFF
{φ} |= ψ iff (φ → ψ) is a tautology;
(iii) (Modus Ponens)
For every Σ, φ, ψ we have that
Σ ∪ {φ, (φ → ψ)} |= ψ.
(iv) (Monotonicity)
If Σ ⊆ Σ1 ⊆ WFF,
φ ∈ WFF and Σ |= φ then also Σ1 |= φ.
(v) (Consequence)
Σ |= (φ → ψ) iff Σ ∪ {φ} |= ψ.
7s-2014
16
Sets and Logic (234293) WS 2014/5
Lecture 7
Deciding logical consequence
(and validity)
In the following we sketch a semantic decision procedure for
the logical consequence.
It is called semantic,
because it resorts to the truth tables
associated with the formulas involved.
A syntactic decision procedure is a
decision procedure which uses as its only data
the formulas themselves.
Syntactic decision procedures will be discussed in Lecture 8
7s-2014
17
Sets and Logic (234293) WS 2014/5
Lecture 7
Theorem 15 (3.2.11)
Semantic decision procedure (algorithm)
for logical consequences
Let Σ be a finite set of well formed formulas in WFF
and φ ∈ WFF.
There is a decision procedure which decides whether Σ |= φ.
Proof:
First we observe that Σ |= φ iff for every
Σ ∪ {¬φ} is not satisfiable.
Let Σ = {ψ1 , . . . , ψn }
Let T T be the truth table for
((
^
ψi ) ∧ ¬φ)
i=1,...,n
This truth table is well defined and finite.
Σ |= φ iff T T is constant with unique value 0.
Q.E.D.
Back to outline of Lecture 7
7s-2014
18
Sets and Logic (234293) WS 2014/5
Lecture 7
Cost of the decision procedure.
Let the Σ and φ have variables pi0 , pi1 , . . . , pin .
The truth tables has 2n+1 values.
P
Let the k = rk(φ) + ψ∈Σ rk(ψ).
Computing a line of the truth table uses at most 2k boolean
operations.
We have k ≥ n.
So we get at least 2n+1 many boolean operations.
CAN WE DO BETTER ?
7s-2014
19
Sets and Logic (234293) WS 2014/5
Lecture 7
Homework
Generalize the notions
tautology,
logical equivalence,
logical consequence,
truth table associated with a formula, and
functional completeness
to WFFS with semantics given by arbitrary truth tables and find examples
and counter examples for these generalized notions.
7s-2014
20
Sets and Logic (234293) WS 2014/5
Lecture 7
Homework (difficult)
Let k ∈ N.
We define WFF(k) to be the set of formulas φ ∈ WFF
containing only the variables p0 , p1 , . . . , pk .
(i) Count the number of distinct formulas in WFF(k).
(ii) Count the number of formulas in WFF(k) which are pairwise logically
not equivalent.
(iii) How long is the longest sequence of formulas φ0 , φ1 , . . . , φn ∈ WFF(k)
such that for every i < n we have that φi |= φi+1 but φi+1 6|= φi .
Back to outline of Lecture 7
7s-2014
21
Sets and Logic (234293) WS 2014/5
Lecture 7
Definition 16 (3.2.14)
Substitution of variables by formulas
Let φ ∈ WFF be a well formed formula.
A function s : V ar → WFF is called a substitution function.
We define inductively a function
subst : WFF × WFFV ar → WFF.
subst(φ, s) is the formula
obtained from φ and s by replacing
all the variables pi in φ simultaneously by s(pi ).
Basis: subst(pi , s) = s(pi ),
subst(F, s) = F, subst(T, s) = T.
Closure: If φ, φ1 , φ2 ∈ WFF then
subst((φ1 ∧ φ2 ), s) = (subst(φ1 , s) ∧ subst(φ2 , s));
subst((φ1 ∨ φ2 ), s) = (subst(φ1 , s) ∨ subst(φ2 , s));
subst((φ1 → φ2 ), s) = (subst(φ1 , s) → subst(φ2 , s));
subst(¬φ, s) = ¬subst(φ, s).
7s-2014
22
Sets and Logic (234293) WS 2014/5
Lecture 7
Examples 17
Let s be given by:
s(p1 ) = p3 , s(p2 ) = (p1 ∧ p5 ),
s(p3 ) = (φ1 ∨ p5 ), . . .
Strictly speaking s(p3 ) = (φ1 ∨ p5 ) is defined only for specific ψ ∈ WFF,
but subst is also well defined for this usage.
• subst(p2 , s) = (p1 ∧ p5 ).
subst(p3 , s) = (φ1 ∧ p5 ).
• subst(((p1 → p2 ) ∨ p3 ), s) =
((p3 → (p1 ∧ p5 )) ∨ (φ1 ∨ p5 ))
• Make your own examples for s, φ
and compute subst(φ, s).
To compute subst((p4 ∧ p1 ), s)
we would have to know the value of s(p4 ).
7s-2014
23
Sets and Logic (234293) WS 2014/5
Lecture 7
Proposition-Exercise 18 (3.2.16)
Finite dependency of substitution
Let φ ∈ WFF be a formula with all its propositional variables
in the set {p0, p1, ..., pn}.
Let s1 and s2 be two substitution functions
such that for every i ≤ n
s1(pi) = s2(pi).
Then
subst(φ, s1) = subst(φ, s2).
7s-2014
24
Sets and Logic (234293) WS 2014/5
Lecture 7
Proposition-Exercise 19 (3.2.17)
(i) Let φ ∈ WFF be not satisfiable and s be a substitution
function. Then subst(φ, s) is not satisfiable.
(ii) Let φ ∈ WFF be a tautology and s be a substitution function.
Then subst(φ, s) is a tautology.
(iii) Let φ ∈ WFF, z ∈ Ass an assignment
and s a substitution function.
Define the assignment z 0 ∈ Ass by z 0(pi) = MP L(s(pi), z).
Then MP L(φ, z 0) = MP L(subst(φ, s), z).
7s-2014
25
Sets and Logic (234293) WS 2014/5
Lecture 7
Homework
• Define substtree : WFTF → WFTF.
• Recall the definitions of the functions
write : WFTF → WFF and
read : WFF → WFTF.
Show that
read(subst(φ, s)) = substtree (read(φ), s)
and
write(substtree (T, s)) = subst(write(T ), s)
• Visualize for yourself the effect of
substitution on WFTF.
Back to outline of Lecture 7
7s-2014
26
Sets and Logic (234293) WS 2014/5
Lecture 7
Normal Forms
This subsection introduces several normal forms of well formed
formulas . In general, a normal form of a formula φ ∈ WFF is a
formula ψ ∈ WFF which is equivalent to φ and whose syntax is
constraint by certain limitations such as
(i) negation symbols are only permitted
if they occur immediately before a variable
(negational normal form NNF);
(ii) when building a formula,
conjunctions are applied last
(conjunctive normal form CNF), or
(iii) when building a formula,
disjunctions are applied last
(disjunctive normal form DNF).
7s-2014
27
Sets and Logic (234293) WS 2014/5
Lecture 7
Definition 20 (3.2.19)
Negational Normal Form
We define a subset NNF ⊆ WFF inductively as follows:
Basis:
The atomic formulas (variables) pi are in NNF.
¬pi ∈ NNF.
F, T ∈ NNF.
Closure:
If φ, ψ ∈ NNF so are (φ ∧ ψ) and (φ ∨ ψ).
Remarks:
Note that formulas in NNF do not contain the symbol →.
As (φ → ψ) is logically equivalent to (¬φ ∨ ψ),
formulas containing → ‘somehow’ contain a ‘hidden’ negation.
7s-2014
28
Sets and Logic (234293) WS 2014/5
Lecture 7
Examples 21
In NNF:
p1, F, (p1 ∧ p2), (p7 ∨ p0)
((p1 ∧ ¬p2) ∨ (¬p2 ∨ F))
Not in NNF:
((p1 ∧ ¬p2) ∨ (¬p2 → F))
((p1 ∧ ¬p2) ∨ ¬(¬p2 ∨ F))
¬¬p1, ¬T
7s-2014
29
Sets and Logic (234293) WS 2014/5
Lecture 7
Theorem 22 (3.2.22)
Negational Normal Form Theorem
For every formula φ ∈ WFF
there is a formula ψ ∈ NNF such that:
(i) φ is equivalent to ψ and
(ii) φ and ψ have the same variables.
7s-2014
30
Sets and Logic (234293) WS 2014/5
Lecture 7
Proof idea
We define a procedure (function)
mvin : WFTF → WFTF
inductively.
The procedure transforms the
tree presentation
of a formula by moving the negations to the leaves while preserving logical
equivalence.
Let Tφ be the tree obtained from φ by the Unique Readability Theorem.
By abuse of notation we shall write nevertheless φ instead of Tφ .
7s-2014
31
Sets and Logic (234293) WS 2014/5
Lecture 7
Proof of Theorem 22, I
The procedure mvin
Basis:
(i) mvin(pi ) = pi ,
mvin(F) = F,
mvin(T) = T.
(ii) mvin(¬pi ) = ¬pi ,
mvin(¬F) = T,
mvin(¬T) = F.
7s-2014
32
Sets and Logic (234293) WS 2014/5
Lecture 7
Proof of Theorem 22, II
The procedure mvin
Closure:
(i) mvin(¬¬φ) = mvin(φ);
(ii) mvin((φ1 ∧ φ2 )) = (mvin(φ1 ) ∧ mvin(φ2 ));
(iii) mvin((φ1 ∨ φ2 )) = (mvin(φ1 ) ∨ mvin(φ2 ));
(iv) mvin((φ1 → φ2 )) = (mvin(¬φ1 ) ∨ mvin(φ2 ));
(v) mvin(¬(φ1 ∧ φ2 )) = (mvin(¬φ1 ) ∨ mvin(¬φ2 ));
(vi) mvin(¬(φ1 ∨ φ2 )) = (mvin(¬φ1 ) ∧ mvin(¬φ2 ));
7s-2014
33
Sets and Logic (234293) WS 2014/5
Lecture 7
Proof of Theorem 22, III
The procedure mvin
As mvin is defined for WFTF rather then WFF
it is well defined.
It is also easy to verify (cf. Proposition 10)
that logical equivalence is preserved.
In other words:
Claim:
mvin(φ) is logically equivalent to φ.
7s-2014
34
Sets and Logic (234293) WS 2014/5
Lecture 7
Proof of Theorem 22, IV
Applying the procedure to get NNF
Given a formula φ ∈ WFTF and a node a we define the subformula φa to be
the formula described by the tree below a.
• If every subformula of φ is in NNF, so is φ.
• Otherwise iterate the following:
– Choose the left most upper most node a labeled with ¬.
The subformula φa of φ is not in NNF unless it is a leave.
– Compute mvin(φa ) = ψ.
– In φ replace φa by ψ.
This replacement preserves logical equivalence.
• The iteration terminates, as the negation at a drops to levels below.
7s-2014
35
Sets and Logic (234293) WS 2014/5
Lecture 7
Homework
In the previous slide we used the phrase
In φ replace φa by ψ.
We want to make this precise:
• Let pnew be a variable which does not occur in φ.
• Let s be the substitition s(pnew ) = φa .
• Let s0 be the substitition s0 (pnew ) = ψ.
• Find a formula θ such that subst(θ, s) = φ.
• With this θ the result of replace is subst(θ, s0 ).
7s-2014
36
Sets and Logic (234293) WS 2014/5
Lecture 7
Definition 23 (3.2.23)
Conjunctive and Disjunctive Normal Form, I
DISJ, CONJ ⊆ WFF:
We define subsets CNF, DNF ⊆ WFF inductively in two stages
as follows.
We first define DISJ, CONJ ⊆ WFF:
Basis:
pi and ¬pi are in DISJ. F, T ∈ DISJ.
pi and ¬pi are in CONJ. F, T ∈ CONJ.
Closure:
If φ, ψ ∈ DISJ so (φ ∨ ψ) ∈ DISJ.
If φ, ψ ∈ CONJ so (φ ∧ ψ) ∈ CONJ.
7s-2014
37
Sets and Logic (234293) WS 2014/5
Lecture 7
Definition 24 (3.2.23, contd)
Conjunctive and Disjunctive Normal Form, II
CNF, DNF ⊆ WFF:
Basis:
DISJ ⊆ CNF. CONJ ⊆ DNF.
Closure:
If φ, ψ ∈ CNF so (φ ∧ ψ) ∈ CNF.
If φ, ψ ∈ DNF so (φ ∨ ψ) ∈ DNF.
7s-2014
38
Sets and Logic (234293) WS 2014/5
Lecture 7
Examples 25
In CNF:
((p1 ∨ ¬p2 ) ∧ (¬p2 ∨ F))
Not in CNF:
((p1 ∨ ¬p2 ) ∧ (¬p2 → F))
((p1 ∧ ¬p2 ) ∨ ¬(¬p2 ∨ F))
In DNF:
(p1 ∨ p2 ), (p1 ∧ p2 )
((p1 ∨ ¬p2 ) ∨ (¬p2 ∨ F))
((p1 ∧ ¬p2 ) ∨ (¬p2 ∨ F))
Not in DNF:
((p1 ∨ ¬p2 ) ∧ (¬p2 ∨ F))
((p1 ∨ ¬p2 ) ∧ (¬p2 → F))
7s-2014
39
Sets and Logic (234293) WS 2014/5
Lecture 7
Theorem 26 (3.2.25)
Conjunctive Normal Form Theorem
For every formula φ ∈ WFF
there is a formula ψ ∈ CNF such that:
(i) φ is equivalent to ψ and
(ii) φ and ψ have the same variables.
7s-2014
40
Sets and Logic (234293) WS 2014/5
Lecture 7
Proof idea
The proof is similar to the proof of the
Negational Normal Form Theorem.
We define inductively a function
mvout : NNF → NNF
whose domain is the set of
tree presentations of formulas in NNF.
This will give usthe tree presentation of the desired formula in
CNF.
7s-2014
41
Sets and Logic (234293) WS 2014/5
Lecture 7
Proof of Theorem 26, I
Definition of mvout
Basis:
If φ ∈ CNF then mvout(φ) = φ.
Closure:
Assume φ, φ1, φ2 are in CNF.
We distinguish two (minimal) cases how a formula in NNF is not in CNF:
(i) mvout((φ ∨ (φ1 ∧ φ2))) = (mvout((φ ∨ φ1)) ∧ mvout((φ ∨ φ2)));
(ii) mvout(((φ1 ∧ φ2) ∨ φ)) = (mvout((φ1 ∨ φ)) ∧ mvout((φ2 ∨ φ)));
The remaining details are left as HOMEWORK.
7s-2014
42
Sets and Logic (234293) WS 2014/5
Lecture 7
Proof of Theorem 26, II
Applying mvout to get CNF
Given a formula φ ∈ WFTF and a node a we define the subformula φa to be
the formula described by the tree below a.
• If every subformula of φ is in CNF, so is φ.
• Otherwise iterate the following:
– Choose the left most upper most node a labeled with ∨, which has a
∧ immediately below it.
The subformula φa of φ is not in CNF.
– Compute mvout(φa ) = ψ.
– In φ replace φa by ψ. This replacement preserves logical equivalence.
• The iteration terminates, as the disjunction at a drops to levels below.
7s-2014
43
Sets and Logic (234293) WS 2014/5
Lecture 7
Functional completeness
of truth tables revisited
Note that the construction in
the proof of the
functional completeness of truth tables
gives for any truth table T T a ψ ∈ CNF.
This can be exploited for an alternative proof of Theorem 26.
7s-2014
44
Sets and Logic (234293) WS 2014/5
Lecture 7
Proposition-Exercise 27 (3.2.27)
Disjunctive Normal Form
For every formula φ ∈ WFF
there is a formula ψ ∈ DNF
such that:
(i) φ is equivalent to ψ and
(ii) φ and ψ have the same variables.
7s-2014
45
Sets and Logic (234293) WS 2014/5
Lecture 7
Computing Σ |= φ
We can check whether
Σ |= φ
with truth tables.
Look at the situation where we have:
Σ |= φ
Σ |= (φ → ψ)
Σ |= ψ
Do we have to compute all the truth tables
to verify all three lines?
7s-2014
46
Sets and Logic (234293) WS 2014/5
Lecture 7
Rules for computing Σ |= φ
Given Σ ⊆ WFF{→,F} we want to compute the set
Con(Σ) = {φ ∈ WFF{→,F} : Σ |= φ}
incrementally.
We shall give an inductive definition of a set Ded(Σ).
The generating sequence which shows that φ ∈ Ded(Σ) is called a
proof sequence.
It is customary to write Σ ` φ for φ ∈ Ded(Σ).
Then we shall show that
Ded(Σ) = Con(Σ).
In other words, the inductively defined set Ded(Σ)
and the set Con(Σ) coincide.
Back to outline of Lecture 7
7s-2014
47
Sets and Logic (234293) WS 2014/5
Lecture 7
The origins of Truth Table Semantics
for Propositional Logic
After R. Zach, Completeness before Post, BSL 5.3 (1999)
Charles Sanders Pierce 1839-1914
Ludwig Wittgenstein 1889-1951
Pierce discusses it already in 1867 but it was widely overlooked. He also was the first to see
the analogy between boolean formulas and circuits.
His work was rediscovered in the 1930ties.
Wittgenstein introduces Truth Table Semantics before 1918
(published in 1921) in his Tractatus Logico-Philosophicus
E. Post (see next slide) independently also introduced Truth Table Semantics.
7s-2014
48
Sets and Logic (234293) WS 2014/5
Lecture 7
The origins of the Completeness Theorem
for Propositional Logic
Emil Post 1897-1954
Paul Bernays 1888-1977
E. Post is usually credited for his complete treatment of Propositional Logic, including
Truth Table Semantics, in his PhD 1920, see his paper Introduction to a general theory of
elementary propositions, American Journal of Mathematics, 1921.
P. Bernays’ work was underestimated because he was extremely modest and mostly known
for having worked out the details of Hilbert’s lectures on logic. His work on Propositional
Logic is his Habilitation from 1918.
Recent analysis of the handwritten notes reveals that Bernays did much more, indeed Paul
Bernays was the Logician behind Hilbert’s Program.
7s-2014
49
Sets and Logic (234293) WS 2014/5
Lecture 7
Boolean Algebra vs Propositional Logic
George Boole 1815-1864
M. H. Stone 1903-1989
G. Boole introduced the laws of Boolean Algebras of sets.
The meaning function assigns subsets of a fixed set to the variables.
M.H. Stone showed that every Boolean Algebra can be presented as an Algebra of Sets.
The connection to Propositional Logic is given by changing the assignment:
Let A be a non-empty set. Now assignements have values in ℘(A).
The empty set plays the role 0, and A plays the role of 1.
Back to outline of Lecture 7
7s-2014
50
Sets and Logic (234293) WS 2014/5
Lecture 7
Why attend lectures?
• If you rely on lectures alone to learn, they will be ineffective.
You can watch lectures on youtube or attend class lectures,
but without study, you won’t learn much.
• You need to study to learn.
That involves reading, writing, and homework assignments.
Also, talk about your object of study with others who are learning it.
• The more active your participation, the better.
Homework, research, projects, labs, reports, writing papers and
computer programs, artwork, creative writing, demonstrations,
and presentations all help more than the
passive participation of reading, watching videos, and attending classes.
• But [do] attend those class lectures.
Use them as your central tool for organizing your learning.
They’re the cadence that keeps you on pace.
Quoted from quora: David Joyce, Always learning
7s-2014
51
© Copyright 2026 Paperzz