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How to represent f() in first order logic so that following scenario holds?

M=f() N=f()

where M $!=$ N

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  • $\begingroup$ You don't. Axioms of equality deman that $M=N$ if $M=f()$ and $N=f()$. $\endgroup$ – Hagen von Eitzen May 19 '17 at 6:25
  • $\begingroup$ how to represent a function f so that $M != N$ $\endgroup$ – Tom May 19 '17 at 6:34
  • $\begingroup$ You cannot: a function in mathematics is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. $\endgroup$ – Mauro ALLEGRANZA May 19 '17 at 6:42
  • $\begingroup$ But the symbol M=f( ) is meaningless: what is the input value ? When you say "a function that return different values each time?" what do you mean with "each time"? different input values? are there instant of time ? $\endgroup$ – Mauro ALLEGRANZA May 19 '17 at 6:43
  • $\begingroup$ Yeah, by definition a function takes exactly one value on each argument. And there is no such notion as "each time" in logic. What you can do however is for instance to consider a function with one more argument, $t$, express that $t$ ranges over a linearly ordered set, and think of $f(x,t)$ as being the value "at time $t$" of a "function that depends on time". $\endgroup$ – Régis May 19 '17 at 7:07

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