#I think my proof is incorrect,and at best too wordy/handwavy. How can I fix this? Please help 🤗

17 messages · Page 1 of 1 (latest)

dull geode
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(For plomble #12): Suppose F ⊆ G. Suppose x ∈ ⋃F. Then by definition of ⋃F,We can let A be an arbitrary set in F,And let x be an arbitrary element in A. Then,because F ⊆ G,A ∈ G. So since x ∈ A,By the definition of ⋃G,x ∈ ⋃G. Thus,we have shown that for an arbitrary element x in ⋃F,x is also in ⋃G,So we can conclude that ⋃F ⊆ ⋃G.

Scratch work:
First,I started by assuming F ⊆ G. Then,analyzing the logical form of the goal I assumed x ∈ ⋃F,which changed my goal to x ∈ ⋃G,meaning I had to find a set in G such that x was in A. I then looked to the given x ∈ ⋃F,which told me there existed a set A in F such that x ∈ A. So by existential instantiation,I could use this given to introduce the set A with x ∈ A. But then since F ⊆ G,A ∈ G,and with x ∈ A,this demonstrates a set A in G such that x ∈ A.
Please,if I botched this up please tear it up and correct my mistakes 👍

This also ties into a more general plomble I have with proofs: I might understand the givens and goal,but when it’s time to actually make the write up I just get stuck at translating it into English. So,optionally if you have advice to help there that’d be appreciated. Thanks

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rigid flintBOT
dull geode
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Is my proof correct? If it’s not,where’d I go wrong and how can it be fixed? Is it too wordy,and is there an alternative way to do it?

forest flower
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Reading your proof, I think your general idea is correct but there are some things that are phrased awkwardly/incorrectly

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In particular this part

Suppose x ∈ ⋃F.
Then by definition of ⋃F, we can let A be an arbitrary set in F, and let x be an arbitrary element in A.

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Since you have already chosen some x in ∪F and you want x∈A, you cannot just let A be an arbitrary element of F. Instead what you are guaranteed is that there exists an A in F such that x∈A, and this is what you need in the proof.

It also does not make sense to let x be an arbitrary element of A because you have already picked out this x.

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I can see what you get the logic correct in your "scratch work"

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The rest of the proof looks OK.

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This is how I would personally write it:

Let x ∈ ∪F. Then x ∈ A for some A ∈ F. Since F ⊂ G, we know A ∈ G also. Therefore x ∈ ∪G.

dull geode
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That part you highlighted tripped me up for a good while 😅

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Suppose F ⊆ G. Let x ∈ ∪F. Then x ∈ A for some A ∈ F. Since F ⊂ G, we know A ∈ G also. Therefore x ∈ ∪G. And because F ⊆ G,A ∈ G. So since x ∈ A,By the definition of ⋃G,x ∈ ⋃G. Thus,we have shown that for an arbitrary element x in ⋃F,x is also in ⋃G,So we can conclude that ⋃F ⊆ ⋃G

How’s this?

forest flower
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Yes it's alright

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A little wordy but that's to be expected at this level

dull geode
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Yes,I have a problem with wordiness,I might “understand” how the givens infer the conclusion from a givens/goal list,but then wording it is my weakness

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Thank you so much for the help,I’ll copy paste it and take a look back at the proof

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.solved