Incorrect Answer.
We consider the body of the program's only while-statement
z = z + a;
a = a + 1;
To see whether
2z = a + 1 
is an invariant for the while-statement,
we place it as a postcondition at the bottom of the loop-body and
push it upwards through that body to determine the weakest
precondition that guarantees
2z = a + 1
to hold after the execution of the loop's body:
2(z + a) = (a + 1) + 1 Weakest Precondition
z = z + a;
2z = (a + 1) + 1 Assignment
a = a + 1;
2z = a + 1 Assignment
- The weakest precondition we determined is
2(z + a) = (a + 1) + 1 , which we may simplify to
2z + a = 2 .
- With that at hand, the postcondition 2z = a + 1 is an invariant for the
program's only while-statement if, and only if,
the weakest precondition is implied by the invariant candidate
2z = a + 1
in conjunction with the condition of the while-statement's boolean
guard a != y+1.
- However, the implication
(a = y + 1) 
(2z = a + 1)
(2z + a = 2) is not true in general.
(Please find values for a, x, y, and z for which this implication is false.)
Back to Question.
Next Question.