\documentclass[12pt]{article}
\usepackage{amsfonts}

\voffset=-3.0cm
\hoffset=-2.6cm
\textwidth=17.5cm
\textheight=24cm

\renewcommand{\not}{\neg}
\renewcommand{\and}{\wedge}
\renewcommand{\or}{\vee}
\newcommand{\impl}{\Rightarrow}
\newcommand{\lequiv}{\Leftrightarrow} %equivalence as logical connective
\newcommand{\all}{\forall}
\newcommand{\exi}{\exists}
\newcommand{\elc}{\models}  % semantic logical consequence
\newcommand{\ylc}{\vdash}  % syntactic logical consequence
%\equiv is already defined for === semantic equivalence
\newcommand{\union}{\cup}
\newcommand{\intersect}{\cap}
\newcommand{\true}{{\mathbb{T}}}
\newcommand{\false}{{\mathbb{F}}}
\newcommand{\nov}[1]{{\overline{#1}}}

\pagestyle{empty}

\begin{document}

{\bf Logic 1, WS 2015.
Homework 6, given Nov 27, due Dec 3.}

\bigskip

\noindent
1. Find the sequent rule for existential assumption.

\bigskip

\noindent
2. Find the sequent rule for existential goal.

\bigskip

\noindent
3. Prove by sequent calculus:
$$(\exists_x P[x]) \impl Q\ \ylc\ \forall_x (P[x] \impl Q) $$

\bigskip

\noindent
4.
Construct the sequent proof of:
$$\exists_x (P[x] \impl Q)\ \ylc\ (\forall_x P[x]) \impl Q$$

\end{document}
