1. **Logical Formulae:**
	1. $Q=\{e\mid e \text{Prop}\},I=Q,F=\{\bot,\top\}$
	2. $$\frac{}{\top\mapsto \top},\frac{}{\bot\mapsto \bot},\frac{e_1\mapsto e_1'}{e_1\land e_2 \mapsto e_1'\land e_2},\frac{}{\top \land e_2 \mapsto e_2},\frac{}{\bot \land e_2 \mapsto \bot},\frac{e\mapsto e'}{\neg e \mapsto \neg e'}$$
2. **Long, but not difficult:**
	1. We do this with induction over $p$. We see that $$\frac{q\stackrel{*}{\mapsto}q,q\stackrel{*}{\mapsto}r}{q\stackrel{*}{\mapsto}r}.$$Assuming that $$\frac{p_1\stackrel{*}{\mapsto}q,q\stackrel{*}{\mapsto}r}{p_1\stackrel{*}{\mapsto}r}.$$ we see that$$\frac{\frac{p\mapsto p_1,p_1\stackrel{*}{\mapsto}q,q\stackrel{*}{\mapsto}r}{p\mapsto p_1,p_1\stackrel{*}{\mapsto}r}}{p\mapsto r}.$$
	2. We need to prove $p(e_1)=\neg e_1 \stackrel{*}{\mapsto}\bot$ for all $e_1 \stackrel{*}{\mapsto} \top$. We do induction of $e_1$. Our induction hypothesis is that $p(e_1')$ is true. We see that $e_1\mapsto \top$ that by our rules $\neq e_1 \mapsto \neg