1. Logics for Data and Knowledge Representation
Exercises: Propositional Logic
Fausto Giunchiglia, Rui Zhang and Vincenzo Maltese

2. Outline Syntax Modeling a problem Semantics Logical implication
Symbols and formation rules
Modeling a problem
Semantics
Assignments, models
Logical implication
Reasoning
Satisfiability and validity
Some nice problems

3. Symbols in PL Which of the following symbols are used in PL?
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Which of the following symbols are used in PL?
⊓  ⊤ ∨ ≡ ⊔ ⊑ → ↔ ⊥ ∧ ⊨
Which of the following symbols are in well formed formulas?

4. Symbols in PL (solution)
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Which of the following symbols are used in PL?
⊓  ⊤ ∨ ≡ ⊔ ⊑ → ↔ ⊥ ∧ ⊨
Which of the following symbols are in well formed formulas?
Remember the BNF grammar:
<Atomic Formula> ::= A | B | ... | P | Q | ... | ⊥ | ⊤
<wff> ::= <Atomic Formula> | ¬<wff> | <wff>∧ <wff> | <wff>∨ <wff> | <wff> → <wff> | <wff> ↔ <wff> 4

5. Formation rules Which of the following is not a wff?
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Which of the following is not a wff?
 MonkeyLow ∨ BananaHigh
  MonkeyLow  BananaHigh
MonkeyLow   BananaHigh
MonkeyLow →  GetBanana
NUM. 3 IS WRONG! 5

6. Modeling: Bananas
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Highlight relevant words, define a PL language and a theory for the problem below.
Bananas may differ in many ways. However, there are red and yellow bananas. I like bananas, but I eat only yellow bananas. If I do not eat at least a banana I get crazy.
CHANGED 6

7. Modeling: Bananas (possible solution)
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Bananas may differ in many ways. However, there are red and yellow bananas. I like bananas, but I eat only yellow bananas. If I do not eat at least a banana I get crazy.
CHANGED
L = {RedBanana, YellowBanana, EatBanana, GetCrazy}
T = {EatBanana → YellowBanana,  EatBanana → GetCrazy} 7

8. Truth valuations and Truth Tables
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
A truth valuation on a PL language L is a mapping ν that assigns to each formula P of L a truth value ν(P).
A truth table is composed of one column for each input variable and one (or more) final column for all of the possible results of the logical operation that the table is meant to represent. Each row of the truth table therefore contains one possible assignment of the input variables, and the result of the operation for those values.
LOGICAL
OPERATION
VARIABLES A B
A∧B T F
POSSIBLE
ASSIGNEMENTS

9. Example Calculate the Truth Table of the following formulas:
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Calculate the Truth Table of the following formulas:
(1) A ∧ B; (2) P ∨ Q; (3) X ↔ Y.
VARIABLES
(1)
(2)
(3) A B
A∧B
A∨B
A↔B T F
POSSIBLE
ASSIGNEMENTS

10. Exercise: Truth Valuations
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Compute truth valuations for the formulas:
(A∨B)∧¬C
(A∨B)→C
(A∨B→C∨D∨E)∧(¬F↔A)∧(¬F∨G∧¬H∨F)∧(¬I→¬(D∧J))∧(¬J∨¬ D∨E)∧F 10

11. Provide the models for the propositions
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
RECALL: a truth valuation ν is a model for a proposition P iff ν(P) = true
List the models for the following formulas:
A ∧ ¬B
(A ∧ B) ∨ (B ∧ C)
A ∨ B → C
¬A ↔ B ↔ C A B
A ∧ ¬ B T F
MODEL 11

12. Entailment
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
RECALL:  ⊨ ψ iff all models satisfying the formulas in  also satisfy ψ
Let A, B, C be propositional sentences.
(a) If A ⊨ B ∧ C, then A ⊨ B or A ⊨ C or both?
(b) What if A ⊨ B ∨ C?
Proof: The only model satisfying ‘B ∧ C’ is ν = {B=T, C=T}.
If ‘A ⊨ B ∧ C’ then ν should be also a model of ‘A’. However, since ν assigns true to B, ν is a model of ‘B’. Similarly, ν is also a model of ‘C’. So, (a) above is true for both.
Assume now that ‘A ⊨ B ∨ C’. A model of ‘B ∨ C’ is ν = {B=T, C=F}.
ν is not a model of ‘C’, therefore A ⊭ C. The other cases are similar.

13. Exercise: prove entailment
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Given that:
P = (A ∨ B ) ∧ (C ∨ D ∨ E)
Q1 = A ∨ B
Q2 = (A ∨ B ∨ C) ∧ ((B ∧ C ∧ D) → E)
Q3 = (A ∨ B) ∧ (D ∨ E)
Does P ⊨ Qi ?
Proof: Let X = A ∨ B, Y = D ∨ E, then we can rewrite:
P = X ∧ (¬C ∨ Y); Q1 = X; Q2 = (X ∨ C) ∧ (¬B ∨ ¬C ∨ Y); Q3 = X ∧ Y
P ⊨ Q1 is obvious.
Since ‘X ⊨ X ∨ C’ and ‘(¬ C ∨ Y) ⊨ (¬B ∨ ¬C ∨ Y)’, then P ⊨ Q2.
Since ‘Y ⊨ (¬C ∨ Y)’, then Q3 ⊨ P (and not vice versa).

14. Exercise: prove entailment using truth tables
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Given that:
P = (A ∨ B ) ∧ (C ∨ D ∨ E)
Q1 = A ∨ B
Q2 = (A ∨ B ∨ C) ∧ (B ∧ C ∧ D → E)
Q3 = (A ∨ B) ∧ (D ∨ E)
(1) List all truth assignments such that P ⊨ Qi
(2) Is there any assignment such that P ⊨ Qi for all i?
Solution to (1): First compute the truth tables for all the propositions above. Then, list all rows for which both P and Qi are true.
Solution to (2): Check whether there is any assignment for which all the sentences above are true. 14

15. Logical implication and deduction
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Deduction (⊢) is the process of inferring new knowledge from known facts.
Logical implication (→) is a way to include deduction directly in the language. It is an alternative way to implement deduction.
For instance: ¬(A ∨ B) → (¬A ∧ ¬B) iff ¬(A ∨ B) ⊢ (¬A ∧ ¬B)
We provided some well known tautologies in the theoretical part (e.g. the De Morgan Law above) 15

16. Prove using truth tables, the following deductions
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Double negative elimination
 P ⊢ P
Conjunction introduction / elimination
(a) {P, Q} ⊢ P∧Q; (b) P ∧ Q ⊢ P; (c) P ∧ Q ⊢ Q.
Disjunction introduction / elimination 
(a) P ⊢P ∨ Q; (a) Q ⊢ P ∨ Q; (c) {P ∨ Q, P → R, Q → R} ⊢ R
Bi-conditional introduction / elimination
(P → Q) ∧ (Q → P) ⊢ (P ↔ Q) 
De Morgan
(a) (P ∧ Q) ⊢ P ∨ Q; (b) (P ∨ Q) ⊢ P ∧ Q 16

17. Proofs of the Deduction Rules
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL P Q
P∧Q
P∨Q
P↔Q T F P Q
¬¬P
P→Q
Q→P
¬(P∧Q)
¬P∨¬Q
¬(P∨Q)
¬P∧¬Q T F 17

18. Soundness and Completeness
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
A deductive system is sound if any sentence P that is derivable from a set Г of sentences is also a logical consequence of that set Г.
A deductive system is complete if every sentence P that is a semantic consequence of a set of sentences Γ can be derived in the deduction system from that set.
A soundness property provides the initial reason for counting a logical system as desirable. The completeness property means that every validity (truth) is provable. 18

19. A = (p → r) ∧ (q → s) ∧ (r ∨ s) →(p ∨ q)
Prove validity (I)
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Suppose p, q, r, s are four propositional sentences, is the following sentence valid?
A = (p → r) ∧ (q → s) ∧ (r ∨ s) →(p ∨ q)
A way to prove validity is to show that the proposition entails ⊤. This can be done by applying well known tautologies (e.g. De Morgan). In alternative we can show using truth tables that all the assignments are true.
(p → r) ∧ (q → s) ∧ (¬r ∨ ¬s) → (¬p ∨ ¬q) =
¬((p → r) ∧ (q → s) ∧ (¬r ∨ ¬s)) ∨ ¬p ∨ ¬q =
(p ∧ ¬r) ∨ (q ∧ ¬s) ∨ (r ∧ s) ∨ ¬p ∨ ¬q =
= ⊤

20. Prove validity (II) 20 p q r s A T F
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL p q r s A T F 20

21. Calculus with Tableaux
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Γ = {A B, B, A}
Γ = {B A, B}
Γ = {(A  B), A} A B A B
B  A
A  B A B A B
A  B
closed
closed A B
closed
closed 21

22. Calculus with Tableaux
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
Γ = {(A B)  (B  A), B} B
(A  B)  ( B  A)
(A  B)
 B  A
A  B B A
closed A B
closed 22

23. Are you ‘Sherlock Holmes’?
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
There was a robbery in which a lot of goods were stolen. The robber(s) left in a truck. It is known that :
(1) Nobody else could have been involved other than A, B and C.
(2) C never commits a crime without A's participation.
(3) B does not know how to drive.
Is A innocent or guilty?
Proof: The 3 points above can be translated in PL as follows:
(1) θ1 = A ∨ B ∨ C;
(2) θ2 = C → A;
(3) θ3 = B → (B ∧ A) ∨ (B ∧ C)
Does {θ1 , θ2 , θ3} ⊨ A ? Yes!
We can prove it by showing that θ1 ∧ θ2 ∧ θ3 → A is a tautology.

24. Knights and Knaves
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie.
You meet two inhabitants: Zoey and Mel.
Zoey tells you that Mel is a knave.
Mel says ‘Neither Zoey nor I are knaves’.
Can you determine what are they? (who is a knight and who is a knave?)
Proof: The two sentences above can be translated in PL as follows:
(1) Z → M; (2) M → Z ∧ M.
We can use truth tables to prove that there are two possible situations:
- Both lie (they are both knaves)
- Zoey tells the true (is a Knight) and Mel lies (is a knave)

25. Recall the DPLL algorithm
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
REVIEW :: REASONING IN PL :: THE DPLL PROCEDURE :: OBSERVTIONS :: CONCLUSIONS
Input: a proposition P in CNF
Output: true if "P satisfiable" or false if "P unsatisfiable"
boolean function DPLL(P) {
if consistent(P) then return true;
if hasEmptyClause(P) then return false;
foreach unit clause C in P do
P = unit-propagate(C, P);
foreach pure-literal L in P do
P = pure-literal-assign(L, P);
L = choose-literal(P);
return DPLL(P  L) OR DPLL(P  L); }
Consistency Check
Empty Close Test
Unit Propagation
Pure Literal Elimination
Splitting rule 25

26. Use DPLL to prove satisfiability
SYNTAX :: MODELING A PROBLEM :: SEMANTICS :: LOGICAL IMPLICATION :: REASONING :: SOME NICE PROBLEMS :: DPLL
B  ¬A  (¬C  A)  (B  C)
1. Is it consistent? YES, there are no contradictions
2. Are there any empty clauses? No, go ahead.
3. Assign the right truth value to all literals and simplify the formula:
Assign B = T, A = F and then the formula simplifies to ¬C
4. Assign the right value to pure literals:
Assign C = F. Done.
5. No need for the splitting rule
The formula is satisfiable at least for the assignment (a model for it):
{A = F, B = T, C = F}