• Steve Dice@sh.itjust.works
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    2 years ago

    No, they’re not.

    Let’s assume they are. Let funky function be defined as:

    int funky() {
        a=0
        b=1
        if ( a==1 ) {
            b=1
        }
        return(a)
    }
    

    Since a==1 if, and only if, b=1, in particular a==1 if b=1. We have b=1, therefore a==1. It follows funky will always return 1 but… it doesn’t. QED.