Is you + sign the same as Or and your * sign the same as And? I believe so. So with this the two "sentences" corresponding to your two logial expressions would be (if I haven't made any mistakes) respectively:
(a and c) or (c and d) or (a and b and d) or (!a and !b and d) or (a
and !b and !d) or (!a and !c and !d)
and
(a or !c or d) and (!a or !b or c or d) and (a or !b or c or !d) and
(!a or b or c or !d)
As Mathematica expressions these are respectively:
expression1=(a && c) || (c && d) || (a && b && d) || ( ! a && ! b && d) || (a && ! b && ! d) || ( ! a && ! c && ! d)
and
expression2= (a || ! c || d) && ( ! a || ! b || c || d) && (a || ! b || c || ! d) && ( ! a || b || c || ! d)
And we can see if these are the sam by using BooleanTable:
In[14]:= BooleanTable[expression1, {a, b, c, d}]
Out[14]= {True, True, True, False, True, True, False, True, True, \
False, False, True, True, False, True, True}
In[15]:= BooleanTable[expression2, {a, b, c, d}]
Out[15]= {True, True, True, False, True, True, False, True, True, \
False, False, True, True, False, True, True}
In[16]:= BooleanTable[expression2, {a, b, c, d}] ===
BooleanTable[expression1, {a, b, c, d}]
Out[16]= True
So, what about Wolfram Alpha? One problem is that forming the And of the two word expressions is too long for the Wolfram Alpha input field... so it goes nowhere.
One possibility is to do two Wolfram Alpha calls
((a and c) or (c and d) or (a and b and d) or (!a and !b and d) or (a
and !b and !d) or (!a and !c and !d)) truth table
and
((a or !c or d) and (!a or !b or c or d) and (a or !b or c or !d) and
(!a or b or c or !d)) truth table
and visually compare the truth tables.
perhaps there are some other approaches in pure Wolfram Alpha that others can suggest.