Basic Diagram Question

Broccoli

Can you solve this or is this invalid

A - (some) - B
B - (all) - c

Tim Hortons

This is a valid argument form because the existential quantifier precedes the universal quantifier. When chained up, it follows that: A ←some→ B → C = A ←some→ C.

A while ago, back when I was CCing, I put together this master list of all the valid and invalid argument forms and the "rules" that accompany each type. I figure you might find it helpful.
(Note: on this list, your question pertains to Valid Form 4)

The VALID Forms (100% Certainty)

Universal Causation
Rule: The only valid inference that can be made from a universal causal relationship is its contrapositive

Form 1: Sufficient/Necessary (A → B + X is A = X is B )
Form 2: Contrapositive of Form 1 ( A → B + X is /B = X is /A)
Form 3: Sufficient/Necessary Extension ( A → B → C = A → C)

Chain Inferences
Rule: The existential quantifier must precede the universal quantifier.

Form 4: Some/All/Some ( A ←some→ B → C = A ←some→ C)
Form 5: Most/All/Most (A ‑most→ B → C = A ‑most→ C)

Same “Sufficient” Variable
Rule: Both premises have to share the same (let’s call it ‘sufficient’) variable.

Form 6: All/All/Some ( A → B + A → C = B ←some→ C)
Form 7: All/Some/Some ( A → B + A ←some→ C = B ←some→ C)
Form 8: All/Most/Some ( A → B + A -most→ C = B ←some→ C)
Form 9: Most/Most/Some ( A -most→ B + A -most→ C = B ←some→ C)

The INVALID Forms (Not 100% Certainty)

Universal Causation
Rule: The only valid inference that can be made from a universal causal relationship is its contrapositive.

Form 1: Reversing i.e. triggering the necessary ( A → B + X is B ≠ X is A)
Form 2: Negating i.e. negating the sufficient ( A → B + X is /A ≠ X is /B)

Inference Chains
Rule: The existential quantifier must precede the universal quantifier, also implying that a UQ has to be present in the chain.

Form 3: All/Some Chain (A → B ←some→ C ≠ A ←some→ C)
Form 4: All/Most Chain (A → B -most→ C ≠ A -most→ C)
Form 5: Some/Some (A ←s→ B ←s→ C ≠ A ←s→ C)
Form 6: Most/Most (A -m→ B -m→ C ≠ A -m→ C or A ←some→ C)

Same “Sufficient” Variable
Rule: Just no.

Form 7: Some/Some (A ←some→ B + A ←some→ C ≠ B ←s→ C

Broccoli

@"Tim Hortons" Wow this is amazing. Thank you so much

Tim Hortons

@Broccoli180 No problem, happy to help!

LCMama2017

@BinghamtonDave I think this is the post you may have seen. Its a great list. I'm going to print it out. Thanks @"tim hortons"

btownsquee

This is valid.

Whenever I'm confused, I think about JY's bucket method.

What if there are a whole bunch of As and just one is B. All Bs are Cs. So that means that one A is also C. Therefore we can validly conclude that A some C.

Broccoli

Thank you. I'll watch bucket method video again.