# Help on proving that there is at most 1 x such that A(x)

I'm looking for someone that can explain this lecture slide to me in a more clear language. I have an exam in a few days and i just cannot wrap my head around why this means that there is only one x s.t. A(x).
In my understanding, you take ∀y for A(y) and then cross-reference ∀x and ∀y to find the value of x and y that both satisfy A(x). However, i don't clearly see what this value of y is, because to me it seems completely arbitrary based on what you choose for y and therefore ∀x∀y((A(x)∧A(y))→ x = y ) does not exclude the possibility of multiple x values that can satisfy A(x) as there can be multiple instances where x=y.
If anyone can give me an example or tips on how to visualize this it would really helpfull!

• NO; assume that there are x1 and x2 such that both satisfy A. This means that A(x1) ∧ A(x2) is true and so from the formula above you will derive - using Modus Ponens - that x1=x2. Jan 21 at 16:38
• "At most" because the formula is true also when there are none. But if there are, they all "coincide" (see argument above). Jan 21 at 16:52
• Check with the natural numbers (starting from 0) and with the property "less than 2". You will see that the formula is not true. Jan 21 at 16:55
• Jan 21 at 17:18
• Exactly...... but not "random" sets, but elements whatever of the same "domain" or universe of discourse. Jan 21 at 17:54