how do I prove {P,Q((P & (Q V R))=(S & T))} |- T if the rule for disjunction introduction is {(P=Q)} |- (~P V Q) ?

Question

how do I prove {P,Q((P & (Q V R))=>(S & T))} |- T if the rule for disjunction introduction is {(P=>Q)} |- (~P V Q) ?

blockcolder · Answer

What does P,Q in the first expression mean?

anonymous · Answer

P, Q, and ((P & (Q V R))=>(S & T)) are the premisses of the argument and t is the conclusion to prove

blockcolder · Answer

Since Q is true, then (Q or R) is true.
Since P and (Q or R) are both true, by modus ponens, (S and T) is true.
Since (S and T) is true, then T is true.

anonymous · Answer

right, but I have to use the rules of logical inference to show that

anonymous · Answer

in the form of a formal proof how do I demonstrate what you just showed? I started like this:
1 1 P Assumption
2 2 Q Assumption
3 3 ((P & (Q V R))=>(S & T)) Assumption
3 4 (~(P & (Q V R)) V ( S & T) 3 disjunction introduction
but now what?

blockcolder · Answer

So you're not allowed to use modus ponens yet?

anonymous · Answer

I can use modus ponens, but how would it help in this case?

anonymous · Answer

I would need to prove ((P & (Q V R)) first

blockcolder · Answer

I'm confused about the numbers in the formal proof. Can you explain those?

anonymous · Answer

the first number represents the line that supports the statement you are making, which is always a premiss or a list of premisses, the second number is just the line number, then the statement, finally the rule you are using to derive the statement and the line which provides immediate justification (any premisses that justify this line must be included in the first line because they by default also are used to support the statement being made)

anonymous · Answer

so I wrote line 4 the way I did because I am using premiss 3 and the rule of disjunction introduction so my justification is in line 3

blockcolder · Answer

It's easy to prove P & (Q V R):
2 5 (Q V R) 2 Disjunctive introduction
1,5 6 P & (Q V R) 1,5 Conjunction introduction
Then modus ponens can be used on the next line.

anonymous · Answer

ah but the problem is my teacher insists that is use a different deffinition for disjunction introduction. His rule is really just a logically equivalent substitution, but I am assuming he has his reasons. His rule is: given ( p => q) we infer (~p V q) so I cannot use the method you mentioned, even though it makes the most sense...