
There are some conditions:
A. input strings(s, t) were valid Anagram.
B. x ∈ string(s), y ∈ string(t) ∑ (x)i == ∑ (y)i .
C. x ∈ string(s), y ∈ string(t) Π (x)i == Π (y)i . 
And a statement:
D. set1{x1,y1}, set2{x2,y2}; x1,x2,y1,y2 ∈ ['a', +∞ ]; if x1y1 == x2y2, x1+y1 == x2+y2; then set1==set2. 
Conditional relation:
I. A => B
II. A => C
III. B+C <=> A (why?) maybe you are question why 'B+C' is sufficient and necessary conditions of 'A'. my answer is that the Statement 'D' was a True Statement. I will prove it now. 
Prove：
First. We assume Statement 'D' was a False Statement. So There are positive integer 'a' and 'b' made the following equation be established:
x1 = x2 + a ( a != 0);
y1 = y2 + b ( b != 0);
Note: when a =0 and b = 0, set1(x1, y1)=set2(x2, y2).
Second. Let's bring it into known conditions:
x1 + y1 = x2 + y2
=>(x2+a) + (y2+b) = x2 + y2
=> a+b = 0
=> a=bx1y1 = x2y2
=>(x2+a)(y2+b) = x2y2
=>x2y2 + ay2 + bx2 + ab = x2y2
=> ay2 + bx2 +ab = 0
∵ a=b
=> by2 + bx2  b^2 = 0
∵ a != 0 , b != 0
=> y2 + x2  b = 0
=> x2 = y2 + b
∵ y1 = y2 + b
=> y1 = x2
Corollaries:
=> x1 = y2
At last. We Proved Set1(x1, y1)=Set2(x2, y2).
However, Statement 'D' was a True Statement!
bool isAnagram(char* s, char* t) {
int b1 = 0, b2 = 0;
int c1 = 1, c2 = 1;
int len1 = (int)strlen(s);
int len2 = (int)strlen(t);
if(len1 != len2){
return false;
}
int i = 0;
while(*(s+i) != '\0'){
b1 += *(s+i);
b2 += *(t+i);
c1 *= *(s+i);
c2 *= *(t+i);
i++;
}
return (b1==b2)&&(c1==c2);
}