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also ...

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7.1--7.2, *7.3--7.6

\def \ititle {Logic I}

\def \isubtitle {Lecture 08}

\begin{center}

{\Large

\textbf{\ititle}: \isubtitle

}

\iemail %

\end{center}

Readings refer to sections of the course textbook, \emph{Language, Proof and Logic}.

\section{Everything Is Broken}

\emph{Reading:} §9.1, §9.2

Everything is broken: ∀x Broken(x)

Something is broken: ∃x Broken(x)

9.1 odd numbers only

9.2 even numbers only

9.8–-9.10

\section{All Squares Are Blue}

\emph{Reading:} §9.2, §9.3, §9.5

9.4--9.5

9.8--9.10

\section{What does ∀ mean?}

\emph{Reading:} §9.4

We give the meaning of ∀ by specifying what it takes for a sentence containing ∀ to be true:

\begin{enumerate}

\item Give every object a name.

\item For each name in turn, create a new sentence like this: delete the quantifier and replace all instances of the variable it binds with that name.

\item If ALL of the new sentences are true, so is the original sentence.

\end{enumerate}

\section{Vegetarians Are Evil}

\emph{Reading:} §9.2, §9.3, §9.5

## Counterexamples with Quantifiers

\section{Counterexamples with Quantifiers}

| Evil(x) | HatesMeat(x) | Vegetarian(x) |

Ayesha (a) | no | ?no | yes |

HatesMeat(a) → Vegetarian(a)

logic support

S0.08

Monday, 08 February 2016

19:00-20:00

I recently compiled all the rules of proofs on a sheet and now use this sheet as a constant reference while working on them. It has improved the way i work substancially so that logic no longer makes me want to die inside ... as much.

## We knew that it was invalid ...

Hi, we have been working on the logic exercises ... ended in a 15 minute argument between two maths students and two philosophy students

The exercise is 5.6 which says:

'From P ∧ Q and ¬P, infer R'

We knew that it was invalid because of not being able to infer R when it is not involved at all in the premise,

An argument is **logically valid **just if there’s no possible situation in which the premises are true and the conclusion false.

however the problem arose when we were discussing the premises.

Is it a contradiction to say that ... that P and Q, then have a second statement that says it's not the case that P?

And if so, then if the premises are contradictory does that automatically make the argument invalid by default?

\section{DeMorgan: ¬(A ∧ B) ⫤⊨ ¬A ∨ ¬B}

\emph{Reading:} §3.6, §4.2

`$\leftmodels\models$' means `is logically equivalent to', so for now `has the same truth table as'.

A $\leftmodels\models$ ¬¬A

¬(A ∧ B) $\leftmodels\models$ (¬A ∨ ¬B)

¬(A ∨ B) $\leftmodels\models$ (¬A ∧ ¬B)

A → B $\leftmodels\models$ ¬A ∨ B

¬(A → B) $\leftmodels\models$ ¬(¬A ∨ B) $\leftmodels\models$ A ∧ ¬B

Here's a useful equivalence: double negations cancel out (at least in logic).

A ⫤⊨ ¬¬A

is logically equivalent to

i.e. has the same truth table as

¬(A ∧ B) | ⫤⊨ | (¬A ∨ ¬B) |

¬(A ∨ B) | ⫤⊨ | (¬A ∧ ¬B) |

A → B | ⫤⊨ | ¬A ∨ B |

¬(A → B) | ⫤⊨ | ¬(¬A ∨ B) | ⫤⊨ | A ∧ ¬B |

A | B | A ∧ B | ¬(A ∧ B) | ¬A | ¬B | ¬A ∨ ¬B |

T | T | T | F | F | F | F |

T | F | F | T | F | T | T |

F | T | F | T | T | F | T |

F | F | F | T | T | T | T |

A | B | ¬(A ∨ B) | ¬A ∧ ¬B |

T | T | F | F |

T | F | F | F |

F | T | F | F |

F | F | T | T |

A | B | A → B | ¬A ∨ B |

T | T | T | T |

T | F | F | F |

F | T | T | T |

F | F | T | T |

A | B | A → B | ¬(A → B) | ¬(¬A ∨ B) | A ∧ ¬B |

T | T | T | F | F | F |

T | F | F | T | T | T |

F | T | T | F | F | F |

F | F | T | F | F | F |

3.19

4.15--18

7.1--7.2, *7.3--7.6

3.19

4.31

\usepackage{pdfpages}

\includepdf[pages={1}]{img/rules.pdf}