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- - **Am I doing this right? (Predicate logic)**
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Am I doing this right? (Predicate logic)Hey thanks for checking the thread! So i have to prove a conclusion is true with predicate logic I don't have a step by step solution so i tried myself That's my problem : Conclusion : ∃x : ¬Q(x) Premises : 1.∀ : x (P(x) → ∀y : (Q(x) → R(x,y)))2 ∃x : (P(x) ∧ ∃y : ¬R(x,y))3. ∃x ∃y [ P(x) ∧ ¬R(x,y)] (pulled out quantifiers) 4. ∃x [ P(x) ∧ ¬R(x,b)] EI(3), not sure if it's right 5. P(a) ∧ ¬R(a,b) EI(4) 6. P(a) Simp(5) 7. ¬R(a,b) Simp(5) 8. ∀x ∀y [ P(x) → Q(x) → R(x,y) ] pulled out quantifiers 9. ∀x [ P(x) → Q(x) → R(x,b) ] UI [8] 10. P(a) → Q(a) → R(a,b) UI[9] 11. All three parts have implication between them therefore i can switch order of operations like this : ( P(a) → Q(a) ) → R(a,b) 12. ( ¬ P(a) ∨ Q(a) ) → R(a,b) solved the inner parentheses with Impl1 rule 13. ¬ ( ¬ P(a) ∨ Q(a)) ∨ R(a,b) same rule for outer one 14. ( P(a) ∧ ¬Q(a) ) ∨ R(a,b) De Morgan's + double negation, keeping the second part the same. 15. ( R(a,b) v ¬Q(a) ) ∧ ( R(a,b) v P(a)) distributive law 16. R(a,b) v ¬Q(a) Simp(15) 17. R(ab) v P(a) Simp(15) 18. ¬Q(a) disjunctive syllogism (16,7) 19 ∃x : ¬Q(x) EG[1,8] First, If you had time to follow all this thank you in advance! Now i'd like to know whether this is solved correctly? Did I make aNY mistakeS and if I did, please show me where and point me in the right direction! |

Can anyone please help me with this problem, i'm stuck! Thanks! |

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