3. case x:
x
0;
p (
(p)+1);p
q ((p)+(q)+1);p
q ((p)+(q)+1);p
q ((p)+(q)+1);
(x)+1 
Prop
f:{f:Formula| Q(f) }.
case f:x 0;p ((p)+1);pq ((p)+(q)+1);pq ((p)+(q)+1);pq ((p)+(q)+1);
(x)+1
| 1 | 2. x: {f:Formula| Q(f) } 3. case x: x 0;p ((p)+1);pq ((p)+(q)+1);pq ((p)+(q)+1);pq ((p)+(q)+1);
(x)+1 |
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