 Supp can save modus ponens by denying that the argument is really of that form. “\(A\Rightarrow B\); \(A\); so \(B\)” is demonstrably valid when \(A\) and \(B\) are propositions. For instance, if \(\mathbf{p}(A) = 90\)% and \(\mathbf{p}_A (B) = 90\)% the lowest possible value for \(\mathbf{p}(B)\) is 81%. The “consequent” of (2), “If Reagan doesn’t win, Anderson will win”, is not a proposition. The argument is really of the form “If \(A \amp B\), then \(C\); \(A\); so if \(B\) then \(C\)”. This argument form is invalid (Supp and Stalnaker agree). It is one of the many argument forms which do preserve certainty, but do not preserve high probability. Take the case where \(C = A\), and we have “If \(A \amp B\) then \(A\); \(A\); so if \(B\) then \(A\)”. The first premise is a tautology and falls out as redundant; and we are left with “\(A\); so if \(B\) then \(A\)”. We have already seen that this is invalid: I can think it very likely that Sue is lecturing right now, without thinking that if she was seriously injured on her way to work, she is lecturing right now.