If the proposition ¬p→v is true, then ¬p∨(p→q) ? Please check my solution. 
If the proposition ¬p→v is true, then the truth value of the
  proposition ¬p∨(p→q), where ¬ is negation, ∨ is inclusive OR and → is
  implication, is
  
  
*
  
*True
  
*False
  
*Multiple Values
  
*Cannot be determined
  


I try to explain

Given ,¬p→v = p+v is valid
now ,  proposition ¬p∨(p→q) = ¬p+q ,
hence using rule of inference ,
= (p+v)→(¬p+q) = ¬p.¬v + ¬p + q = ¬p+q ,
so it cannot be determined.
Edit : Is my method correct ?
 A: The formula $¬p→v$ is True in two cases :

either when $p$ is True (i.e. $\lnot p$ False) or when $v$ is True.

Thus, the assumption does not force a truth-value for $p$.
Consider now $¬p∨(p→q)$; we have three cases :
(i) $q$ True; in this case $p→q$ is True, irrespective of the truth-value of $p$, and so $¬p∨(p→q)$ is T.
(ii) $q$ False; now two subcases :


*

*(ii-a) $p$ True; in this case $p→q$ is False, and so $¬p∨(p→q)$ is F.

*(ii-b) $p$ False; in this case $¬p∨(p→q)$ is T.

Conclusion : you are right, cannot be determined.
Note : please, note that the previous explanation is nothing more than the description in words of the truth-table entered for $p,q,v$, observing that in all the six rows with T for the first formula, the second one has different values.

Regarding your solution : your transformations are correct, but what are they aimed to ? 
The $v$ in $\lnot p \to v = p \lor v$ will not disappear, and thus you will not be able to reach the second formula.
But this "impossibility" (quite evident) is not easy to prove that way.
A: I don't see what $v$ has to do with the question. It seems that the statement is true if $\neg p$, and false if $p \land \neg q$.
