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Can I do a conjunction in the antecedent of a conditional? i.e. step 7 in my derivation below legit?

  1. $A \rightarrow B\qquad\qquad$ (Premise)
  2. $A \rightarrow C\qquad\qquad$ (Premise)
  3. $A \phantom{{}\rightarrow C}\qquad\quad\quad$ (Premise)
  4. $B \phantom{{}\rightarrow C}\qquad\quad\quad$ (By 1 and 3, modus ponens)
  5. $C \phantom{{}\rightarrow C}\qquad\quad\quad$ (By 2 and 3, modus ponens)
  6. $A \wedge C\quad\quad\quad\qquad$ (By 3 and 5, conjunction)
  7. $(A \wedge C)\rightarrow B\quad\quad$ (By 1 and conjunction of the antecedent)
  8. Therefore $C \rightarrow B\quad$ (Conclusion) (Simplification of conjunction in step 7)

Is the following reasoning logically sound, to explain step 7?

a. $(A\wedge C)\rightarrow A\qquad\qquad$ (Simplification of conjunction in 6)
b. $A\rightarrow B\qquad\qquad\qquad$(Premise)
c. $(A\wedge C)\rightarrow A\rightarrow B\qquad$ (By a and b)
d. Therefore, $(A\wedge C)\rightarrow B\quad$ (Hypothetical syllogism)

Greatly appreciate any advice. Thanks!

< 10 Dec >

Hi everybody, many thanks for your inputs. After a good night's sleep, here I am at it again:

A: ε --> 0
B: f(x) --> L
C: δ --> 0

  1. A --> B (Premise 1)
  2. A --> C (Premise 2)
  3. A (Premise 3)
  4. B (1,3 MP)
  5. C (2,3 MP)
  6. A ⋀ C (3,5 conjunction)
  7. (A ⋀ C) --> A (6, simplification)
  8. (A ⋀ C) --> B (7,1 hypothetical syllogism)
  9. A --> (C --> B) (8, exportation)
  10. Therefore, C --> B (3,9 MP)
    i.e. as δ --> 0, f(x) --> L

    Problem seems solved in the interim.
    Just need to convince myself that I can allow Premise 2 i.e. as ε --> 0, δ --> 0 (and vice versa, since I think the function that links ε and δ is bijective)
Wei
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2 Answers2

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Note that exactly what rules you may use depend on what system you have defined and what have been proven for that system.

Yes, step 7 is correct. However step 8 is not as it stands. Step 8 does not follow from step 7 but rather it follows , in the same manner as we got step 7, from step 4 and step 5.

In the second part. step c. looks very weird, since you do not have parenteses. However you should be able to go directly from a and b to d, and skip step c. in any case I dont think it helps.

Ove Ahlman
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  • Yes, you're right! Step 8 doesn't follow from 7. Like I mentioned in my reply to BrianO, I'm desperate to prove C --> B, and desperate people tend to do funny things! Pardon me :( I'm going to have to rethink and rework all my premises again. – Wei Dec 09 '15 at 15:12
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Are you trying to derive $C\to B$ from $A\to B, A\to C$? That's not valid. (Take $C = True, A=B=False$.) You never discharge the premise $A$ of step 3.

Your final four-step, a)-d) deduction is valid but for reasons other than the ones you give. How does c) follow from a) and b)? I understand the following derivation, however:

a) $(A\land C)\to A \qquad\text{simplification of conjunction}$

b) $A\to B \qquad\qquad\text{ premise}$

c) $A \qquad\qquad\qquad\text{from 6. and a) by MP}$

d) $B \qquad\qquad\qquad\text{from b) and c) by MP}$

e) $(A\land C) \to B \qquad\text{ from d): from $p$, infer $q\to p$ (weakening)}$.

BrianO
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  • Thanks! I've been pushing the my luck too hard ... context is I'm trying to get a handle on the epsilon-delta proof for limits. But the calculus textbooks I have don't spell out the logical derivation process explicitly. My proposition C is "delta approaches zero", B is "f(x) approaches L". And I'm somewhat desperate to proof that as delta approaches zero, f(x) approaches L - which is the gist of the epsilon-delta definition. – Wei Dec 09 '15 at 15:09
  • Well, $\delta = \delta(\varepsilon)$ is or can be thought of a function of $\varepsilon$, so it's really "as $\varepsilon$ approaches $0$, $f(x)$ approaches $L$". For continuity at $x$: the image under $f$ of the $\delta(\varepsilon)$ "ball" around $x$ is contained in the $\varepsilon$ ball around $L$. In the real line, the "$r$-ball around $x$" is the set of $y$ with $\lvert y-x\rvert < r$. Alias $(x-r, x+r)$. You can't reallly express this using just propositional logic (quantifiers are required). – BrianO Dec 09 '15 at 15:18
  • Thanks again! Back to the drawing board! I suspect I left out an important premise: (ε --> 0) <==> (δ -->0). But just a gut feeling at the moment. Not entirely sure if I can justify this premise, and especially that δ and ε correlates positively. Btw, am reading James Stewart's Calculus textbook. – Wei Dec 09 '15 at 15:43
  • I don't know that calculus book, but I'm sure the definition given is standard. By the way, regarding $\delta(\varepsilon)$ as a function of $\varepsilon$, nothing in the continuity definition suggests that $\delta(\varepsilon)\to 0$ as $\varepsilon\to 0$. For a constant function $f$, you could have $\delta(\varepsilon) = 17$ for all $\varepsilon$. For most functions, $\delta(\varepsilon)$ will go to $0$ as $\varepsilon$ does, but it's not required or implied. – BrianO Dec 10 '15 at 03:41
  • Thanks! I'm learning a lot from you. Appreciate it! – Wei Dec 10 '15 at 05:54
  • You're welcome, and thank you :) – BrianO Dec 10 '15 at 06:13