这是Hoare logic方面的一道题。你能根据题目和具体的解题思路，为我详细讲解解题步骤吗？当我是从来没有接触过这个领域的小白。---```text ans = ans + (i*i*i); {ans = i^2 (i + 1)^2 /4} //Invariant {ans = i^2 (i + 1)^2 /4 ^ i==n} //Invariant and Not(Guard) {ans = n^2 (n + 1)^2 /4} //Postcondition Step 4: Simplification (A) Establishing Invariant: {0 = 0^2 (0 + 1)^2 /4} <<<->>> {0 == 0} <<<->>> {True} (B) Preservation of Invariant: {ans + (i+1)^3 = (i+1)^2 (i + 2)^2 /4} <<<->>> {ans = i^2 (i + 1)^2 /4 } Therefore we have {ans = i^2 (i + 1)^2 /4 ^ i != n} //Invariant and Guard -> {ans = i^2 (i + 1)^2 /4 } which is true by conjunction rules Our overall proof is: {true} //Precondition -> {0 == 0}// by consequence rule {0 = 0^2 (0 + 1)^2 /4} //by assign & compose int i = 0; {0 = i^2 (i + 1)^2 /4} //by assign & compose int ans = 0; {ans = i^2 (i + 1)^2 /4} //Invariant while (i != n) {ans = i^2 (i + 1)^2 /4 ^ i != n} //Invariant and Guard -> {ans = i^2 (i + 1)^2 /4} // by consequence rule -> {ans + (i+1)^3 = (i+1)^2 (i + 2)^2 /4} //by assign & compose i = i + 1; {ans + (i*i*i) = i^2 (i + 1)^2 /4} // by assign ans = ans + (i*i*i); {ans = i^2 (i + 1)^2 /4} //Invariant {ans = i^2 (i + 1)^2 /4 ^ i==n} //Invariant and Not(Guard) {ans = n^2 (n + 1)^2 /4} //Postcondition ```