introduction to logic syntax and ... - the logic...

INTRODUCTION TO LOGIC

Syntax and Semantics ofPropositional Logic

Volker Halbach

Logic is the beginning of wisdom.�omas Aquinas

Introduction

In what follows I look at some formal languages that aremuchsimpler than English and de�ne validity of arguments, ‘truthunder an interpretation’, consistency etc. for these formallanguages.

In logic one abstracts from all stylistic variants etc of naturallanguage and retains just the basic skeleton of the language in aregimented form.

1.6 Syntax vs. Semantics

When presenting a formal language, I proceed in the followingorder:

1 I specify the syntax or grammar of the language; inparticular I de�ne what the sentences of the language are.

2 I specify the semantics of the language; in particular, I saywhat it means for a sentence to be true under aninterpretation (or in a ‘structure’). Once the notion of aninterpretation (or structure) is clear, I can de�ne validity ofarguments etc as for English.


Syntax is all about expressions: words and sentences.

Examples of syntactic claims‘Bertrand Russell’ is a proper noun.‘likes logic’ is a verb phrase.‘Bertrand Russell likes logic’ is a sentence.Combining a proper noun and a verb phrase in this wayyields a sentence.


Semantics is all aboutmeanings of expressions.

Examples of semantic claims‘Bertrand Russell’ refers to a British philosopher.‘Bertrand Russell’ refers to Bertrand Russell.‘likes logic’ expresses a property Russell has.‘Bertrand Russell likes logic’ is true.

2.1 Quotation

Note our use of quotes to talk about expressions.

‘Bertrand Russell’ refers to Bertrand Russell.

Mention�e �rst occurrence of ‘Bertrand Russell’ is an example ofmention.

�is occurrence (with quotes) refers to an expression.

Use�e second occurrence of ‘Bertrand Russell’ is an example ofuse.

�is occurrence (without quotes) refers to a man.

2.1 Quotation



Mention�e �rst occurrence of ‘Bertrand Russell’ is an example ofmention.�is occurrence (with quotes) refers to an expression.

Use�e second occurrence of ‘Bertrand Russell’ is an example ofuse.

�is occurrence (without quotes) refers to a man.

2.1 Quotation



Mention�e �rst occurrence of ‘Bertrand Russell’ is an example ofmention.�is occurrence (with quotes) refers to an expression.

Use�e second occurrence of ‘Bertrand Russell’ is an example ofuse.�is occurrence (without quotes) refers to a man.

2.2 The Syntax of the Language of Propositional Logic

Syntax: English vs. L.

English hasmany di�erent sorts of expressions.

Some expressions of English(1) Sentences: ‘Bertrand Russell likes logic’, ‘Philosophers likeconceptual analysis’, etc..

(2) Connectives: ‘it is not the case that’, ‘and’, etc..(3) Noun phrases: ‘Bertrand Russell’, ‘Philosophers’, etc..(4) Verb phrases: ‘likes logic’, ‘like conceptual analysis’, etc..(5) Also: nouns, verbs, pronouns, etc., etc., etc..

L has just two sorts of basic expressions.

Some basic expressions of L(1) Sentence letters: e.g. ‘P’, ‘Q’.(2) Connectives: e.g. ‘¬’, ‘∧’.�ere are also brackets: ‘(’ and ‘)’.




Some expressions of English

(1) Sentences: ‘Bertrand Russell likes logic’, ‘Philosophers likeconceptual analysis’, etc..








(2) Connectives: ‘it is not the case that’, ‘and’, etc..

(3) Noun phrases: ‘Bertrand Russell’, ‘Philosophers’, etc..(4) Verb phrases: ‘likes logic’, ‘like conceptual analysis’, etc..(5) Also: nouns, verbs, pronouns, etc., etc., etc..









Some basic expressions of L

(1) Sentence letters: e.g. ‘P’, ‘Q’.(2) Connectives: e.g. ‘¬’, ‘∧’.�ere are also brackets: ‘(’ and ‘)’.







Some basic expressions of L(1) Sentence letters: e.g. ‘P’, ‘Q’.

(2) Connectives: e.g. ‘¬’, ‘∧’.�ere are also brackets: ‘(’ and ‘)’.


Combining sentences and connectives makes new sentences.

Some complex sentences‘It is not the case that’ and ‘Bertrand Russell likes logic’ make:‘It is not the case that Bertrand Russell likes logic’.

‘¬’ and ‘P’ make: ‘¬P’.‘Bertrand Russell likes logic’ and ‘and’ and ‘Philosophers like conceptualanalysis’ make:‘Bertrand Russell likes logic and philosophers like conceptual analysis’.

‘P’, ‘∧’ and ‘Q’ make: ‘(P ∧ Q)’.

Logic convention: no quotes around L-expressions.

P, ∧ and Q make: (P ∧ Q).



Some complex sentences

‘It is not the case that’ and ‘Bertrand Russell likes logic’ make:‘It is not the case that Bertrand Russell likes logic’.


‘P’, ‘∧’ and ‘Q’ make: ‘(P ∧ Q)’.







‘P’, ‘∧’ and ‘Q’ make: ‘(P ∧ Q)’.






‘¬’ and ‘P’ make: ‘¬P’.

‘Bertrand Russell likes logic’ and ‘and’ and ‘Philosophers like conceptualanalysis’ make:‘Bertrand Russell likes logic and philosophers like conceptual analysis’.

‘P’, ‘∧’ and ‘Q’ make: ‘(P ∧ Q)’.







‘P’, ‘∧’ and ‘Q’ make: ‘(P ∧ Q)’.




Connectives

Here’s the full list of L-connectives.

name in English symbolconjunction and ∧

disjunction or ∨

negation it is not the ¬

case thatarrow if . . . then →

double arrow if and only if ↔


�e syntax of L

Here’s the o�cial de�nition of L-sentence.

De�nition(i) All sentence letters are sentences of L:

P,Q , R, P,Q, R, P,Q, R, P, . . .(ii) If ϕ and ψ are sentences of L, then so are:

¬ϕ(ϕ ∧ ψ)(ϕ ∨ ψ)(ϕ → ψ)(ϕ↔ ψ)

(iii) Nothing else is a sentence of L.

Greek letters: ϕ (‘PHI’) and ψ (‘PSI’): not part of L. 35


�e syntax of L


De�nition

(i) All sentence letters are sentences of L:P,Q , R, P,Q, R, P,Q, R, P, . . .

(ii) If ϕ and ψ are sentences of L, then so are:¬ϕ(ϕ ∧ ψ)(ϕ ∨ ψ)(ϕ → ψ)(ϕ↔ ψ)




�e syntax of L



P,Q , R, P,Q, R, P,Q, R, P, . . .

(ii) If ϕ and ψ are sentences of L, then so are:¬ϕ(ϕ ∧ ψ)(ϕ ∨ ψ)(ϕ → ψ)(ϕ↔ ψ)




�e syntax of L




¬ϕ

(ϕ ∧ ψ)(ϕ ∨ ψ)(ϕ → ψ)(ϕ↔ ψ)




�e syntax of L




¬ϕ(ϕ ∧ ψ)

(ϕ ∨ ψ)(ϕ → ψ)(ϕ↔ ψ)




�e syntax of L




¬ϕ(ϕ ∧ ψ)(ϕ ∨ ψ)

(ϕ → ψ)(ϕ↔ ψ)




�e syntax of L




¬ϕ(ϕ ∧ ψ)(ϕ ∨ ψ)(ϕ → ψ)

(ϕ↔ ψ)(iii) Nothing else is a sentence of L.



�e syntax of L




¬ϕ(ϕ ∧ ψ)(ϕ ∨ ψ)(ϕ → ψ)(ϕ↔ ψ)




How to build a sentence of L

Example�e following is a sentence of L:

¬¬(((P ∧ Q)→ (P ∨ ¬R))↔ ¬((P ∨ R) ∨ R))

De�nition of L-sentences (repeated from previous page)

(i) All sentence letters are sentences of L.(ii) If ϕ and ψ are sentences of L, then ¬ϕ, (ϕ ∧ ψ), (ϕ ∨ ψ),(ϕ → ψ) and (ϕ↔ ψ) are sentences of L.





¬¬(((

P

∧ Q)→ (P ∨ ¬R))↔ ¬((P ∨ R) ∨ R))







¬¬(((

P

∧

Q

)→ (P ∨ ¬R))↔ ¬((P ∨ R) ∨ R))







¬¬((

(P ∧ Q)

→ (P ∨ ¬R))↔ ¬((P ∨ R) ∨ R))







¬¬((

(P ∧ Q)

→ (P ∨ ¬

R

))↔ ¬((P ∨ R) ∨ R))







¬¬((

(P ∧ Q)

→ (P ∨

¬R

))↔ ¬((P ∨ R) ∨ R))







¬¬((

(P ∧ Q)

→ (

P

∨

¬R

))↔ ¬((P ∨ R) ∨ R))







¬¬((

(P ∧ Q)

→

(P ∨ ¬R)

)↔ ¬((P ∨ R) ∨ R))







¬¬(

((P ∧ Q)→ (P ∨ ¬R))

↔ ¬((P ∨ R) ∨ R))







¬¬(

((P ∧ Q)→ (P ∨ ¬R))

↔ ¬((

P

∨ R) ∨ R))







¬¬(

((P ∧ Q)→ (P ∨ ¬R))

↔ ¬((

P

∨

R

) ∨ R))







¬¬(

((P ∧ Q)→ (P ∨ ¬R))

↔ ¬(

(P ∨ R)

∨ R))







¬¬(

((P ∧ Q)→ (P ∨ ¬R))

↔ ¬(

(P ∨ R)

∨

R

))







¬¬(

((P ∧ Q)→ (P ∨ ¬R))

↔ ¬

((P ∨ R) ∨ R)

)







¬¬(

((P ∧ Q)→ (P ∨ ¬R))

↔

¬((P ∨ R) ∨ R)

)







¬¬

(((P ∧ Q)→ (P ∨ ¬R))↔ ¬((P ∨ R) ∨ R))







¬

¬(((P ∧ Q)→ (P ∨ ¬R))↔ ¬((P ∨ R) ∨ R))







¬¬(((P ∧ Q)→ (P ∨ ¬R))↔ ¬((P ∨ R) ∨ R))





I mentioned that ϕ and ψ are not part of L.

¬P is a L-sentence.

¬ϕ describes many L-sentences

(but is not one itself).e.g. ¬P, ¬(Q ∨ R), ¬(P↔ (Q ∨ R)), . . .

ϕ and ψ are part of the metalanguage, not the object one.

Object language�e object language is the one we are theorising about.

�e object language is L.

Metalanguage�e metalanguage is the one we are theorising in.

�e metalanguage is (augmented) English.

ϕ and ψ are used as variables in the metalanguage:in order to generalise about sentences of the object language.



¬P is a L-sentence.¬ϕ describes many L-sentences

(but is not one itself).e.g. ¬P, ¬(Q ∨ R), ¬(P↔ (Q ∨ R)), . . .









¬P is a L-sentence.¬ϕ describes many L-sentences

(but is not one itself).

e.g. ¬P, ¬(Q ∨ R), ¬(P↔ (Q ∨ R)), . . .









¬P is a L-sentence.¬ϕ describes many L-sentences (but is not one itself).e.g. ¬P, ¬(Q ∨ R), ¬(P↔ (Q ∨ R)), . . .







2.3 Rules for Dropping Brackets

Bracketing conventions

�ere are conventions for dropping brackets in L.

Some are similar to rules used for + and × in arithmetic.

Example in arithmetic

+ ×

does not abbreviate ( + ) × .

× ‘binds more strongly’ than +.

+ × abbreviates + ( × ).

Conventions in L

∧ and ∨ bind more strongly than→ and↔.

(P → Q ∧ R) abbreviates (P → (Q ∧ R)).

One may drop outer brackets.

P ∧ (Q → ¬P) abbreviates (P ∧ (Q → ¬P)).

One may drop brackets on strings of ∧s or ∨s that arebracketed to the le�.

(

P ∧ Q ∧ R

)

abbreviates ((P ∧ Q) ∧ R). 25


Bracketing conventions�ere are conventions for dropping brackets in L.Some are similar to rules used for + and × in arithmetic.

Example in arithmetic

+ ×

does not abbreviate ( + ) × .



Conventions in L






(

P ∧ Q ∧ R

)




Example in arithmetic + ×

does not abbreviate ( + ) × .× ‘binds more strongly’ than +.


Conventions in L






(

P ∧ Q ∧ R

)




Example in arithmetic + × does not abbreviate ( + ) × .



Conventions in L






(

P ∧ Q ∧ R

)




Example in arithmetic + × does not abbreviate ( + ) × .× ‘binds more strongly’ than +.


Conventions in L






(

P ∧ Q ∧ R

)




Example in arithmetic + × does not abbreviate ( + ) × .× ‘binds more strongly’ than +. + × abbreviates + ( × ).

Conventions in L






(

P ∧ Q ∧ R

)





Conventions in L∧ and ∨ bind more strongly than→ and↔.

(P → Q ∧ R) abbreviates (P → (Q ∧ R)).One may drop outer brackets.



(

P ∧ Q ∧ R

)





Conventions in L∧ and ∨ bind more strongly than→ and↔.(P → Q ∧ R) abbreviates (P → (Q ∧ R)).




(

P ∧ Q ∧ R

)





Conventions in L∧ and ∨ bind more strongly than→ and↔.(P → Q ∧ R) abbreviates (P → (Q ∧ R)).One may drop outer brackets.

P ∧ (Q → ¬P) abbreviates (P ∧ (Q → ¬P)).One may drop brackets on strings of ∧s or ∨s that arebracketed to the le�.

(

P ∧ Q ∧ R

)





Conventions in L∧ and ∨ bind more strongly than→ and↔.(P → Q ∧ R) abbreviates (P → (Q ∧ R)).One may drop outer brackets.P ∧ (Q → ¬P) abbreviates (P ∧ (Q → ¬P)).


(

P ∧ Q ∧ R

)





Conventions in L∧ and ∨ bind more strongly than→ and↔.(P → Q ∧ R) abbreviates (P → (Q ∧ R)).One may drop outer brackets.P ∧ (Q → ¬P) abbreviates (P ∧ (Q → ¬P)).One may drop brackets on strings of ∧s or ∨s that arebracketed to the le�.

(

P ∧ Q ∧ R

)





Conventions in L∧ and ∨ bind more strongly than→ and↔.(P → Q ∧ R) abbreviates (P → (Q ∧ R)).One may drop outer brackets.P ∧ (Q → ¬P) abbreviates (P ∧ (Q → ¬P)).One may drop brackets on strings of ∧s or ∨s that arebracketed to the le�.(P ∧ Q ∧ R) abbreviates ((P ∧ Q) ∧ R). 25




Conventions in L∧ and ∨ bind more strongly than→ and↔.(P → Q ∧ R) abbreviates (P → (Q ∧ R)).One may drop outer brackets.P ∧ (Q → ¬P) abbreviates (P ∧ (Q → ¬P)).One may drop brackets on strings of ∧s or ∨s that arebracketed to the le�.

(

P ∧ Q ∧ R

)


2.4 The Semantics of Propositional Logic

Semantics

Recall the characterisation of validity from week 1.

CharacterisationAn argument is logically valid if and only if there is no interpretation ofsubject-speci�c expressions under which:

(i) the premisses are all true, and(ii) the conclusion is false.

We’ll adapt this characterisation to L.Logical expressions: ¬,∧,∨,→ and↔.Subject-speci�c expressions: P,Q , R, . . .Interpretation: L-structure.


Semantics




We’ll adapt this characterisation to L.

Logical expressions: ¬,∧,∨,→ and↔.Subject-speci�c expressions: P,Q , R, . . .Interpretation: L-structure.


Semantics




We’ll adapt this characterisation to L.Logical expressions: ¬,∧,∨,→ and↔.

Subject-speci�c expressions: P,Q , R, . . .Interpretation: L-structure.


Semantics




We’ll adapt this characterisation to L.Logical expressions: ¬,∧,∨,→ and↔.Subject-speci�c expressions: P,Q , R, . . .

Interpretation: L-structure.


Semantics




We’ll adapt this characterisation to L.Logical expressions: ¬,∧,∨,→ and↔.Subject-speci�c expressions: P,Q , R, . . .Interpretation: L-structure.


L-structures

We interpret sentence letters by assigning them truth-values:either T for True or F for False.

De�nitionAn L-structure is an assignment of exactly one truth-value (T orF) to every sentence letter of L.

ExamplesWe can think of an L-structure as an in�nite list that provides avalue T or F for every sentence letter.

P Q R P Q R P Q R ⋯

A ∶ T F F F T F T T F ⋯

B ∶ F F F F F F F F F ⋯

We useA, B, etc. to stand for L-structures.


L-structures

We interpret sentence letters by assigning them truth-values:

either T for True or F for False.








L-structures

We interpret sentence letters by assigning them truth-values:either T for True or F for False.








Truth-values of complex sentences /L-structures only directly specify truth-values for P, Q, R, . . .

�e logical connectives have �xed meanings.�ese determine the truth-values of complex sentences.Notation: ∣ϕ∣A is the truth-value of ϕ underA.For all L-structuresA and sentences ϕ we have either∣ϕ∣A = T or ∣ϕ∣A = F.

Truth-conditions for ¬�e meaning of ¬ is summarised in its truth table.

ϕ ¬ϕT FF T

In words: ∣¬ϕ∣A = T if and only if ∣ϕ∣A = F.



�e logical connectives have �xed meanings.

�ese determine the truth-values of complex sentences.Notation: ∣ϕ∣A is the truth-value of ϕ underA.For all L-structuresA and sentences ϕ we have either∣ϕ∣A = T or ∣ϕ∣A = F.


ϕ ¬ϕT FF T




�e logical connectives have �xed meanings.�ese determine the truth-values of complex sentences.

Notation: ∣ϕ∣A is the truth-value of ϕ underA.For all L-structuresA and sentences ϕ we have either∣ϕ∣A = T or ∣ϕ∣A = F.


ϕ ¬ϕT FF T




�e logical connectives have �xed meanings.�ese determine the truth-values of complex sentences.Notation: ∣ϕ∣A is the truth-value of ϕ underA.

For all L-structuresA and sentences ϕ we have either∣ϕ∣A = T or ∣ϕ∣A = F.


ϕ ¬ϕT FF T




�e logical connectives have �xed meanings.�ese determine the truth-values of complex sentences.Notation: ∣ϕ∣A is the truth-value of ϕ underA.For all L-structuresA and sentences ϕ we have either∣ϕ∣A = T or ∣ϕ∣A = F.


ϕ ¬ϕT FF T



Worked example

∣ϕ∣A is the truth-value of ϕ underA.ϕ ¬ϕT FF T

Compute the following truth-values.Let the structureA be partially speci�ed as follows.


T F F F T F T T F ⋯

Compute:∣P∣A =

T

∣Q∣A =

F

∣R∣A =

F

∣¬P∣A =

F

∣¬Q∣A =

T

∣¬R∣A =

T

∣¬¬P∣A =

T

∣¬¬Q∣A =

F

∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A =

F

∣R∣A =

F

∣¬P∣A =

F

∣¬Q∣A =

T

∣¬R∣A =

T

∣¬¬P∣A =

T

∣¬¬Q∣A =

F

∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A = F ∣R∣A =

F

∣¬P∣A =

F

∣¬Q∣A =

T

∣¬R∣A =

T

∣¬¬P∣A =

T

∣¬¬Q∣A =

F

∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A = F ∣R∣A = F∣¬P∣A =

F

∣¬Q∣A =

T

∣¬R∣A =

T

∣¬¬P∣A =

T

∣¬¬Q∣A =

F

∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A = F ∣R∣A = F∣¬P∣A = F ∣¬Q∣A =

T

∣¬R∣A =

T

∣¬¬P∣A =

T

∣¬¬Q∣A =

F

∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A = F ∣R∣A = F∣¬P∣A = F ∣¬Q∣A = T ∣¬R∣A =

T

∣¬¬P∣A =

T

∣¬¬Q∣A =

F

∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A = F ∣R∣A = F∣¬P∣A = F ∣¬Q∣A = T ∣¬R∣A = T∣¬¬P∣A =

T

∣¬¬Q∣A =

F

∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A = F ∣R∣A = F∣¬P∣A = F ∣¬Q∣A = T ∣¬R∣A = T∣¬¬P∣A = T ∣¬¬Q∣A =

F

∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A = F ∣R∣A = F∣¬P∣A = F ∣¬Q∣A = T ∣¬R∣A = T∣¬¬P∣A = T ∣¬¬Q∣A = F ∣¬¬R∣A =

F


Worked example





Compute:∣P∣A = T ∣Q∣A = F ∣R∣A = F∣¬P∣A = F ∣¬Q∣A = T ∣¬R∣A = T∣¬¬P∣A = T ∣¬¬Q∣A = F ∣¬¬R∣A = F


Truth-values of complex sentences /

Truth-conditions for ∧ and ∨�e meanings of ∧ and ∨ are given by the truth tables:

ϕ ψ (ϕ ∧ ψ)T T TT F FF T FF F F

ϕ ψ (ϕ ∨ ψ)T T TT F TF T TF F F

∣(ϕ ∧ ψ)∣A = T if and only if ∣ϕ∣A = T and ∣ψ∣A = T.

∣(ϕ ∨ ψ)∣A = T if and only if ∣ϕ∣A = T or ∣ψ∣A = T (or both).

15



Truth-conditions for ∧ and ∨�e meanings of ∧ and ∨ are given by the truth tables:

ϕ ψ (ϕ ∧ ψ)T T TT F FF T FF F F

ϕ ψ (ϕ ∨ ψ)T T TT F TF T TF F F

∣(ϕ ∧ ψ)∣A = T if and only if ∣ϕ∣A = T and ∣ψ∣A = T.∣(ϕ ∨ ψ)∣A = T if and only if ∣ϕ∣A = T or ∣ψ∣A = T (or both).

15



Truth-conditions for→ and↔�e meanings of→ and↔ are given by the truth tables:

ϕ ψ (ϕ → ψ)T T TT F FF T TF F T

ϕ ψ (ϕ↔ ψ)T T TT F FF T FF F T

∣(ϕ → ψ)∣A = T if and only if ∣ϕ∣A = F or ∣ψ∣A = T.

∣(ϕ↔ ψ)∣A = T if and only if ∣ϕ∣A = ∣ψ∣A.



Truth-conditions for→ and↔�e meanings of→ and↔ are given by the truth tables:

ϕ ψ (ϕ → ψ)T T TT F FF T TF F T

ϕ ψ (ϕ↔ ψ)T T TT F FF T FF F T

∣(ϕ → ψ)∣A = T if and only if ∣ϕ∣A = F or ∣ψ∣A = T.∣(ϕ↔ ψ)∣A = T if and only if ∣ϕ∣A = ∣ψ∣A.


Worked example

Let ∣P∣B = T and ∣Q∣B = F.

Compute ∣¬(P → Q)→ (P ∧ Q)∣BWhat is the truth value of ¬(P → Q)→ (P ∧ Q) under B?

1 ∣(P → Q)∣B = F and ∣(P ∧ Q)∣B = F2 ∣¬(P → Q)∣B = T3 ∣¬(P → Q)→ (P ∧ Q)∣B = F

ϕ ¬ϕT FF T

ϕ ψ (ϕ ∧ ψ) (ϕ → ψ)T T T TT F F FF T F TF F F T


Worked example



1 ∣(P → Q)∣B

= F and ∣(P ∧ Q)∣B = F2 ∣¬(P → Q)∣B = T3 ∣¬(P → Q)→ (P ∧ Q)∣B = F

ϕ ¬ϕT FF T



Worked example



1 ∣(P → Q)∣B = F

and ∣(P ∧ Q)∣B = F2 ∣¬(P → Q)∣B = T3 ∣¬(P → Q)→ (P ∧ Q)∣B = F

ϕ ¬ϕT FF T



Worked example



1 ∣(P → Q)∣B = F and ∣(P ∧ Q)∣B

= F2 ∣¬(P → Q)∣B = T3 ∣¬(P → Q)→ (P ∧ Q)∣B = F

ϕ ¬ϕT FF T



Worked example



1 ∣(P → Q)∣B = F and ∣(P ∧ Q)∣B = F

2 ∣¬(P → Q)∣B = T3 ∣¬(P → Q)→ (P ∧ Q)∣B = F

ϕ ¬ϕT FF T



Worked example



1 ∣(P → Q)∣B = F and ∣(P ∧ Q)∣B = F2 ∣¬(P → Q)∣B

= T3 ∣¬(P → Q)→ (P ∧ Q)∣B = F

ϕ ¬ϕT FF T



Worked example



1 ∣(P → Q)∣B = F and ∣(P ∧ Q)∣B = F2 ∣¬(P → Q)∣B = T

3 ∣¬(P → Q)→ (P ∧ Q)∣B = F

ϕ ¬ϕT FF T



Worked example



1 ∣(P → Q)∣B = F and ∣(P ∧ Q)∣B = F2 ∣¬(P → Q)∣B = T3 ∣¬(P → Q)→ (P ∧ Q)∣B

= F

ϕ ¬ϕT FF T



Worked example



1 ∣(P → Q)∣B = F and ∣(P ∧ Q)∣B = F2 ∣¬(P → Q)∣B = T3 ∣¬(P → Q)→ (P ∧ Q)∣B = F

ϕ ¬ϕT FF T



For actual calculations it’s usually better to use tables.

Suppose ∣P∣B = T and ∣Q∣B = F.

Compute ∣¬(P → Q)→ (P ∧ Q)∣BP Q ¬ (P → Q) → (P ∧ Q)

T F T T F F F T F F

ϕ ¬ϕT FF T





Compute ∣¬(P → Q)→ (P ∧ Q)∣BP Q ¬ (P → Q) → (P ∧ Q)T F

T T F F F T F F

ϕ ¬ϕT FF T






T

T

F F F T F F

ϕ ¬ϕT FF T






T

T

F

F

F T F F

ϕ ¬ϕT FF T






T

T

F

F

F

T

F F

ϕ ¬ϕT FF T






T

T

F

F

F

T

F

F

ϕ ¬ϕT FF T






T

T

F

F

F

T F F

ϕ ¬ϕT FF T






T

T F F

F

T F F

ϕ ¬ϕT FF T





Compute ∣¬(P → Q)→ (P ∧ Q)∣BP Q ¬ (P → Q) → (P ∧ Q)T F T T F F

F

T F F

ϕ ¬ϕT FF T





Compute ∣¬(P → Q)→ (P ∧ Q)∣BP Q ¬ (P → Q) → (P ∧ Q)T F T T F F F T F F

ϕ ¬ϕT FF T



Using the same technique we can �ll in the full truth table for¬(P → Q)→ (P ∧ Q)

P Q ¬ (P → Q) → (P ∧ Q)T T

F T T T T T T T

T F

T T F F F T F F

F T

F F T T T F F T

F F

F F T F T F F F

�e main column (in boldface) gives the truth-value of the wholesentence.

ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T T T T T T

T F

T T F F F T F F

F T

F F T T T F F T

F F

F F T F T F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T T T T T T

T F

T

T

F F F T F F

F T

F F T T T F F T

F F

F F T F T F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T T T T T T

T F

T

T

F F F T F F

F T

F

F

T T T F F T

F F

F F T F T F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T T T T T T

T F

T

T

F F F T F F

F T

F

F

T T T F F T

F F

F

F

T F T F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T T T

T

T T

T F

T

T

F F F T F F

F T

F

F

T T T F F T

F F

F

F

T F T F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T T T

T

T T

T F

T

T

F F F

T

F F

F T

F

F

T T T F F T

F F

F

F

T F T F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T T T

T

T T

T F

T

T

F F F

T

F F

F T

F

F

T T T

F

F T

F F

F

F

T F T F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T T T

T

T T

T F

T

T

F F F

T

F F

F T

F

F

T T T

F

F T

F F

F

F

T F T

F

F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T

T

T

T

T T

T F

T

T

F F F

T

F F

F T

F

F

T T T

F

F T

F F

F

F

T F T

F

F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T

T

T

T

T T

T F

T

T

F

F

F

T

F F

F T

F

F

T T T

F

F T

F F

F

F

T F T

F

F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T

T

T

T

T T

T F

T

T

F

F

F

T

F F

F T

F

F

T

T

T

F

F T

F F

F

F

T F T

F

F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T

T

T

T

T T

T F

T

T

F

F

F

T

F F

F T

F

F

T

T

T

F

F T

F F

F

F

T

F

T

F

F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T

T

T

T

T

TT F

T

T

F

F

F

T

F F

F T

F

F

T

T

T

F

F T

F F

F

F

T

F

T

F

F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T

T

T

T

T

TT F

T

T

F

F

F

T

F

FF T

F

F

T

T

T

F

F T

F F

F

F

T

F

T

F

F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T

T

T

T

T

TT F

T

T

F

F

F

T

F

FF T

F

F

T

T

T

F

F

TF F

F

F

T

F

T

F

F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T

T

T

T

T

T

TT F

T

T

F

F

F

T

F

FF T

F

F

T

T

T

F

F

TF F

F

F

T

F

T

F

F

F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T T T

T

T

T

TT F

T

T

F

F

F

T

F

FF T

F

F

T

T

T

F

F

TF F

F

F

T

F

T

F

F

F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T T T

T

T

T

TT F

T

T F F

F

T

F

FF T

F

F

T

T

T

F

F

TF F

F

F

T

F

T

F

F

F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T T T

T

T

T

TT F

T

T F F

F

T

F

FF T

F

F T T

T

F

F

TF F

F

F

T

F

T

F

F

F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T T T

T

T

T

TT F

T

T F F

F

T

F

FF T

F

F T T

T

F

F

TF F

F

F T F

T

F

F

F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T T T

T

T T TT F

T

T F F

F

T

F

FF T

F

F T T

T

F

F

TF F

F

F T F

T

F

F

F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T T T

T

T T TT F

T

T F F

F

T F FF T

F

F T T

T

F

F

TF F

F

F T F

T

F

F

F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T T T

T

T T TT F

T

T F F

F

T F FF T

F

F T T

T

F F TF F

F

F T F

T

F

F

F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T

F

T T T

T

T T TT F

T

T F F

F

T F FF T

F

F T T

T

F F TF F

F

F T F

T

F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T F T T T

T

T T TT F

T

T F F

F

T F FF T

F

F T T

T

F F TF F

F

F T F

T

F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T F T T T

T

T T TT F T T F F

F

T F FF T

F

F T T

T

F F TF F

F

F T F

T

F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T F T T T

T

T T TT F T T F F

F

T F FF T F F T T

T

F F TF F

F

F T F

T

F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T F T T T

T

T T TT F T T F F

F

T F FF T F F T T

T

F F TF F F F T F

T

F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T F T T T T T T TT F T T F F

F

T F FF T F F T T

T

F F TF F F F T F

T

F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T F T T T T T T TT F T T F F F T F FF T F F T T

T

F F TF F F F T F

T

F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T F T T T T T T TT F T T F F F T F FF T F F T T T F F TF F F F T F

T

F F F


ϕ ¬ϕT FF T




P Q ¬ (P → Q) → (P ∧ Q)T T F T T T T T T TT F T T F F F T F FF T F F T T T F F TF F F F T F T F F F


ϕ ¬ϕT FF T



Validity

Let Γ be a set of sentences of L and ϕ a sentence of L.

De�nition�e argument with all sentences in Γ as premisses and ϕ asconclusion is valid if and only if there is no L-structure underwhich:(i) all sentences in Γ are true; and(ii) ϕ is false.

Notation: when this argument is valid we write Γ ⊧ ϕ.

{P → ¬Q ,Q} ⊧ ¬P means that the argument whose premises areP → ¬Q and Q, and whose conclusion is ¬P is valid.Also written: P → ¬Q ,Q ⊧ ¬P

5


Worked example 3

We can use truth-tables to show that L-arguments are valid.

ExampleShow that {P → ¬Q ,Q} ⊧ ¬P.

P Q P → ¬ Q Q ¬ P

�

T T T F F T T F T

�

T F T T T F F F T

�

F T F T F T T T F

�

F F F T T F F T F

Rows correspond to interpretations.One needs to check that there is no row in which all thepremisses are assigned T and the conclusion is assigned F.


Worked example 3



P Q P → ¬ Q Q ¬ P

�

T T T F F T T F T

�

T F T T T F F F T

�

F T F T F T T T F

�

F F F T T F F T F

Rows correspond to interpretations.

One needs to check that there is no row in which all thepremisses are assigned T and the conclusion is assigned F.


Worked example 3



P Q P → ¬ Q Q ¬ P

�

T T T F F T T F T

�

T F T T T F F F T

�

F T F T F T T T F

�

F F F T T F F T F



Worked example 3



P Q P → ¬ Q Q ¬ P� T T T F F T T F T

�

T F T T T F F F T

�

F T F T F T T T F

�

F F F T T F F T F



Worked example 3



P Q P → ¬ Q Q ¬ P

�

T T T F F T T F T� T F T T T F F F T

�

F T F T F T T T F

�

F F F T T F F T F



Worked example 3



P Q P → ¬ Q Q ¬ P

�

T T T F F T T F T

�

T F T T T F F F T� F T F T F T T T F

�

F F F T T F F T F



Worked example 3



P Q P → ¬ Q Q ¬ P

�

T T T F F T T F T

�

T F T T T F F F T

�

F T F T F T T T F� F F F T T F F T F



Other logical notions

De�nitionA sentence ϕ of L is logically true (a tautology) i�:

ϕ is true under all L-structures.

e.g. P ∨ ¬P, and P → P are tautologies.

Truth tables of tautologiesEvery row in the main column is a T.

P P ∨ ¬ P P → PT T T F T T T TF F T T F F T F


De�nitionA sentence ϕ of L is a contradiction i�:

ϕ is not true under any L-structure.

e.g. P ∧ ¬P, and ¬(P → P) are contradictions.

Truth tables of contradictionsEvery row in the main column is an F.

P P ∧ ¬ P ¬ (P → P)T T F F T F T T TF F F T F F F T F


De�nitionSentences ϕ and ψ are logically equivalent i�:

ϕ and ψ are true in exactly the same L-structures.

P and ¬¬P are logically equivalent.P ∧ Q and ¬(¬P ∨ ¬Q) are logically equivalent.

Truth tables of logical equivalents�e truth-values in the main columns agree.

P Q P ∧ Q ¬ (¬ P ∨ ¬ Q)T T T T T T F T F F TT F T F F F F T T T FF T F F T F T F T F TF F F F F F T F T T F


Worked example 4

ExampleShow that the sentence (P → (¬Q ∧ R)) ∨ P is a tautology.

Method 1: Full truth tableWrite out the truth table for (P → (¬Q ∧ R)) ∨ P.Check there’s a T in every row of the main column.

P Q R (P → (¬ Q ∧ R)) ∨ PT T T T F F T F T T TT T F T F F T F F T TT F T T T T F T T T TT F F T F T F F F T TF T T F T F T F T T FF T F F T F T F F T FF F T F T T F T T T FF F F F T T F F F T F


Worked example 4 (cont.)Show that the sentence (P → (¬Q ∧ R)) ∨ P is a tautology.

Method 2: Backwards truth table.Put an F in the main column.Work backwards to show this leads to a contradiction.

P Q R (P → (¬ Q ∧ R)) ∨ P

T F F F

ϕ ¬ϕT FF T

ϕ ψ (ϕ ∧ ψ) (ϕ ∨ ψ) (ϕ → ψ)T T T T TT F F T FF T F T TF F F F T




P Q R (P → (¬ Q ∧ R)) ∨ P

T F

F

F

ϕ ¬ϕT FF T





P Q R (P → (¬ Q ∧ R)) ∨ P

T

F F

F

ϕ ¬ϕT FF T





P Q R (P → (¬ Q ∧ R)) ∨ P

T

F F F

ϕ ¬ϕT FF T





P Q R (P → (¬ Q ∧ R)) ∨ PT F F F

ϕ ¬ϕT FF T





P Q R (P → (¬ Q ∧ R)) ∨ P? F F F

ϕ ¬ϕT FF T


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