SOLUTION: Please help! I can't for the life of me figure this one out!
Proof by natural deduction- Predicate Logic. Use a direct proof to show that the following argument is valid.
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-> SOLUTION: Please help! I can't for the life of me figure this one out!
Proof by natural deduction- Predicate Logic. Use a direct proof to show that the following argument is valid.
Pre
Log On
Question 1135774: Please help! I can't for the life of me figure this one out!
Proof by natural deduction- Predicate Logic. Use a direct proof to show that the following argument is valid.
Premise 1: (∃x)Kx --> (x) (Lx --> Mx)
Premise 2: Kc • Lc
Conclusion: Mc
You can put this solution on YOUR website! Hint : assuming you are using Natural Deduction
From Premise-2 : Kc∧Lc
you have to derive Kc (by ∧-elimination), followed by ∃xKx (by ∃-introduction).
In this way, you can use → -elimination (i.e. modus ponens) with Premise-1 and derive : (∀x)(Lx→Mx).
Now you have to use ∀-elimination (i.e. universal instantiation) with c to get : Lc→Mc.
