ex x st P[x]
proof
L:P[a];
take x=a;
then P[x] by L;
thus thesis;
end;
When demonstrating the existence of x that satisfies P[x], let us assume that P[a] is demonstrated. In this case, it means that for take x=a, x is exists and if we take a, it is demonstrated by L.
(ex x st P[x]) implies for y st Q[y] holds R[y]
proof
assume ex x st P[x];
then consider a such that L:P[a]
......
......
let y such that Q[y];
......
......
thus R[y];
end;
First, we will assume. For assume ex x st P[x];, we use consider and state then consider a such that L:P[a]. This case also needs labels because such that exists. We will leave it like this and to state y st next, we will state let y such that Q[y] to state thus Q[y].
(ex x st P[x]) implies for y st Q[y] holds R[y]
proof
given x such that P[x];
......
......
let y
assume Q[y];
......
......
thus R[y];
end;