Why does this work, finding min value and index for an array

Question

This program takes user input 10 times, then determines the lowest value and the index location for that lowest value.

Why does this if-statement work? The "counter == 0" part, wouldn't that always be false? Meaning the code wouldn't work? Yet it does?

#include <stdio.h>
#pragma warning(disable: 4996)
#define arraySize 10

int getNum(void);

int main()
{
    int myArray[arraySize] = {0};
    int counter = 0;
    
    int minIndex = 0;
    int minValue = 0;
    
    printf("Enter 10 Integer Values\n");
    
    for (counter = 0; counter < arraySize; counter++)
    {
        printf("Enter Integer #%d: ", counter);
        myArray[counter] = getNum();
        
        
        if (counter == 0 || (minValue > myArray[counter]))
        {
            minValue = myArray[counter];
            minIndex = counter;
        }
        
    }
    
    printf("Min: %d, Index: %d.", minValue, minIndex);
    
    return 0;
}



int getNum(void)
{
char record[121] = {0};
int number = 0;
    fgets(record, 121, stdin);
    if( sscanf(record, "%d", &number) != 1 )
    {
        number = -1;
    }
    return number;
}

Answer 1

Lets work with this inputs: 0,1,2,3,4,5,6,7,8,-1

In the first iteration counter==0 is true and minValue > myArray[counter] is false because 0 > 0 return false, the operator || allow to enter the if.

In the following iterations the sentence counter==0 will be false ever, but the minValue > myArray[counter] will be false or true depends on the number. With 1,2,3,4,5,6,7,8 will be false but whit -1 will enter because 0 > -1. And make the min value = -1;

The || operator works like this:

true || true =true
true || false =true
false || true ==true
false || false ==false

To sum up, if there is one true sentence the program will enter in the if sentence.