theorem
  s,kai|= EX H iff ex pai being inf_path of R st pai.0 = s & (pai.1),kai |= H
proof
A1: (ex pai being inf_path of R st pai.0 = s & (pai.1) |= Evaluate(H,kai) )
  implies ex pai being inf_path of R st pai.0 = s & (pai.1),kai|= H
  by Def60;
  s,kai|= EX H iff s|= EX Evaluate(H,kai) by Th7;
  hence thesis by A1,Th14;
end;
