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**HallsofIvy** You **defined** "L(x, y)" so that L must have **two** parameters- "L(x)" is meaningless. You want "∀xP(x)=> L(Ryan, x)".

Seriously? "L(x,y)=> L(x,y)"? "A=> A" is always true for any A! You want just "∀x∀y P(x)^P(y) => L(x,y)".

No. This says "for everyone at the party, there is someone they don't know." That would be impossible if Ryan know everybody. You want "∃x∃y P(x)^P(y) ^ ⌜L(y,x).

Do you not see that this involves **three** people? You want "∀x∀y P(x)^P(y)^ ⌜L(x, y)=> ∃z P(z) ^ L(x, z) ^ L(y, z)".