FULL PROLOG CODE PLEASE Define a recursive predicate factorial$/2,$ that returns the factorial of $($small$)$ integers factorial$($X$,3)$ should return X$=6($i$.$e$.$ the first parameter is the factorial and the second is the integer$)$ factorial$(6,3)$ should return true factorial$(6,$X$)$ should return X$=3($this may or may not work depending on the design of your code$).$ You don't need to ensure this $-$ you can still get $100\%$ without this feature. Discussion: You will need a base case $($a non recusive version of factorial$/2).$ It is: factorial$(1,1).$ The rest of the code required is what to do when the second argument is $>1.$ It is only one line of code.