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Marclan said:iii) ∃z ∀x ∃y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z) → ∀x2 ∃y2 (member(x2) ∧ bicycle(y2) ∧ owns(x2,y2) ∧ brand(y2,z)))
So, can i mark this thread as finally solved?
The qualifiers at the beginning were key. Once you get those in the correct order, it's just:
∃z ∀x ∃y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∧ brand(y,z))