**Maintainer:**admin

Note: 10th was a holiday, 12th was the midterm so yeah works out

*1*A proof¶

From last class:

"All horses are animals. Therefore, all heads of horses are heads of animals."

$$\forall x (Hx \to Ax) \vdash A (\exists x(Hx \land H^*yx) \to \exists x (Ax \land H^*yx))$$

Where $Hx$ means "x is a horse", $H^*xy$ means "x is the head of y", and $Ax$ means "x is an animal".

Strategy for the proof: let's try to get $\forall x ( \quad) \vdash \exists x (\quad) \to \exists x ( \quad )$ using a deduction^{?}.

So working up from the bottom:

$$\forall x(\quad) \vdash \exists x (\quad) \to \exists x (\quad)$$

$$\forall x(\quad), \exists x(\quad) \vdash \exists x (\quad)$$

The last quantifier to be put back on: middle term.

Anyways, whatever. The proof:

Sequent | Justification/rule |
---|---|

(1) $Hx \to Ax,\,Hx \land H^*yx \vdash Ax \land H^*yx$ | Tautology |

(2) $Hx \to Ax,\,Hx \land H^*yx \vdash \exists x (Ax \land H^*yx)$ | $\exists$-introduction from (1) |

(3) $\forall x(Hx \to Ax) \vdash Hx \to Ax$ | $\forall$-elimination |

(4) $\forall x (Hx \to Ax),\,Hx \to Ax,\,Hx \land H^*yx \vdash \exists x (Ax \land H^*yx)$ | Monotonicity from (2) |

(5) $\forall x (Hx \to Ax), \, Hx \land H^*yx \vdash Hx \to Ax$ | Monotonicity from (3) |

(6) $\forall x(Hx \to Ax),\,Hx \land H^*yx \vdash \exists x (Ax \land H^*yx)$ | Transitivity using (4), (5) |

(7) $\forall x (Hx \to Ax),\,\exists x (Hx \land H^*yx) \vdash \exists x(Ax \land H^*yx)$ | $\forall$-elimination |

(8) $\forall x (Hx \to Ax) \vdash \exists x (Hx \land H^*yx) \to \exists x (Ax \land H^*yx)$ | Deduction from (7) |

(9) $\forall x(Hx \to Ax) \vdash \forall y (\exists x(Hx \land H^*yx) \to \exists x(Ax \land H^*yx))$ | $\forall$-introduction from (8) |

*2*Some logical axioms for equality¶

- $\forall x(x = x)$ (reflexivity)
- $\forall x\forall y(x = y \to y = x)$ (symmetry)
- $\forall x \forall y \forall z ((x =y \land y = z) \to x =z)$

These are all properties of equivalence relations, and equality is, in a way, the epitome of equivalence relations.

Since these are logical axioms, you can use them in any line of a proof, in the following manner:

$$\vdash \forall x (x = x)$$

*2.1*Quantifying existence¶

How do you say that there is exactly one boy? (Or Buddha, or whatever.) Well, you'd have to use equality. For instance:

$$\exists x (Bx \land \forall y (By \to x = y))$$

So there is a boy, and anything else that is also a boy is the same as our original boy. So there is only one boy.

How would you say there is at most one boy?

$$\exists x \forall y (By \to y = x)$$

So if there are boys, they are all the same boy.

*2.2*Using the logical axioms¶

Here's a proof that uses logical axiom 1 in its first line:

$$\vdash \exists x (x = x)$$

Sequent | Justification/rule |
---|---|

(1) $\vdash \forall x (x=x)$ | Logical axiom |

(2) $\forall x(x=x) \vdash x = x$ | $\forall$-elimination |

(3) $\vdash x = x$ | Transitivity using (1) and (2) |

(4) $\vdash \exists x (x =x)$ | $\exists$-introduction from (3) |

This fails in an empty universe, but don't worry about that. Loveys doesn't care about empty universes and neither should you.