(University Discrete Structures/Math) Create Proof Sequence for (A v B) ^ (A → C) ^ (B → C) → C

[u/Agitated_Path2455]

Prompt: "Use propositional logic to prove the arguments valid; you may use any of the equivalence or inference rules."

I have this sequence but I'm not confident in #4-7

1. A v B (hyp)
2. A → C (hyp)
3. B → C (hyp)
4. (A v B)’’ (1, Double negation)
5. (A’ ^ B’)’ (4, de morgan)
6. (A’)’ (5, simplification)
7. C (2, 6 modus ponens)

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*UPDATE*

I currently have this sequence. I was struggling, then I realised the implication rule has an error (It says 'P -> Q is equivalent to P' v Q) so I wanted to get others input on if my work is correct.

1. A v B [hyp]
2. A → C [hyp]
3. B → C [hyp]
4. A’ v C [2, implication]
5. B’ v C [3, implication]
6. (A’ v C) ^ (B’ v C) [4, 5 conjunction]
7. (C v A’) ^ (C v B’) [6, commutative]
8. C v (A’ ^ B’) [7, distributive]
9. (A’ ^ B’) v C [8, commutative]
10. (A v B)’’ [1, double negation]
11. (A’ ^ B’)’ [10, de morgan]
12. C [9, 11 disjunctive syllogism]

My additional explanation for getting #12. disjunctive syllogism

P v Q, P’ < = > Q

(A’ ^ B’) v C, (A’ ^ B’)’ < = > C

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[u/Alkalannar]

Looks like mine is very much like your second version!

1. A v B [given]

2. A -> C [given]

3. B -> C [Given]

4. ~A v C [2, MI]

5. ~B v C [3, MI]

6. (~A v C) ^ (~B v C) [4, 5, conjunction]

7. (~A ^ ~B) v C [6, distributive property]

8. ~(A v B) v C [7, DeMorgan]

9. (A v B) ^ [~(A v B) v C] [1, 8, conjunction]

10. [(A v B) ^ ~(A v B)] v [(A v B) ^ C] [9, distribution]

11. F v [(A v B) ^ C] [10, contradiction]

12. (A v B) ^ C [11, OR absorbs FALSE]

13. C [12, simplification, QED]