If we let x=1x=1 and o=–1o=–1, then note that consecutive symbols are replaced by their product. If we consider the product PP of the nine values before and after each operation, we will see that the new PP is the square of the old PP. Hence, PP will always equal 11 after an operation. So nine oo's yielding P=–1P=–1 can never happen.

If we let x=1x=1 and o=–1o=–1, then note that consecutive symbols are replaced by their product. If we consider the product PP of the nine values before and after each operation, we will see that the new PP is the square of the old PP. Hence, PP will always equal 11 after an operation. So nine oo's yielding P=–1P=–1 can never happen.

If we let \(x = 1\) and \(o = – 1\), then note that consecutive symbols are replaced by their product. If we consider the product \(P \) of the nine values before and after each operation, we will see that the new \(P\) is the square of the old \(P\). Hence, \(P \) will always equal \(1 \) after an operation. So nine \(o\)'s yielding \(P = – 1\) can never happen.