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Can anyone help a brainlet?
(a) Prove that the following is a valid deduction rule:
P → Q
Q → R
∴ P → R
(b) Prove that the following is a valid deduction rule for any n ≥ 2:
P1 → P2
P2 → P3
.
.
.
Pn−1 → Pn
∴ P1 → Pn.
"I suggest you don’t go through the trouble of writing out a 2^n
row truth table.
Instead, you should use part (a) and mathematical induction."
Can anyone help with part b. I got part a
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>>8505838
>base case
see (a)
>inductive case
assume it holds for n-1
then
[math]P_1 → P_{n−1} [/math]
[math]P_{n−1} → P_n [/math]
so by (a)
[math] P_1 → P_{n} [/math]
QED
>>
>>8505838
Show it's true for n = 2, then show that it being true for n = k implies that it's true for n = k + 1.
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>>8505860
So I can use the base case with n=3 like in (a)? I thought a base case needed to use the lowest possible n which would be 2.
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>>8505869
i know that...just not sure how to apply it in this context witht he propositional logic
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>>8505877
>Q → R
>∴ Q → R
Wow, so hard.
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