If "P implies (Q or R)" and "Q holds true", what can be said about R?

Question

I encountered the above situation when I was working with these set of equations $$\begin{cases} a^2+bc&=&0 \\ b(a+d)&=&0 \\ c(a+d)&=&0 \\ d^2+bc&=&0 \end{cases}$$ where $a,b,c,d\in \mathbb{C}$.

Suppose $a+d\ne 0$. Then from $(2)$ and $(3)$, we have $b=c=0$. Also from $(1)$ and $(4)$, we have $a^2=0 \Rightarrow a=0$ and $d^2=0 \Rightarrow d=0$. But then $a+d=0$ which contradicts our assumption. Hence, it must be the case that $a+d=0$.

We know that if $b(a+d)=0$ then $b=0$ or $a+d=0$. Since, we showed that $a+d=0$, can $b$ be any arbitrary complex number? Seems intuitive but I cannot seem to show it.

Answer 1

Right, $a+d$ must be $0$. The only thing which is implied by the system is that $b$ and $c$ are complex numbers satisfying $bc = -a^2$. If you consider the equation $b(a+d) = 0$ together with the condition $a+d = 0$, this does not imply anything on $b$.

Answer 2

To satisfy your equations, you can have $a=1$, $d=-1$, $b=z$ and $c=-1/z$ for any non-zero complex number $z$. You can also get $b=0$, by setting $a=b=c=d=0$. So yes, $b$ can be anything.

Answer 3

Note that if $a+d=0$, (1) and (4) reduce to the same and (2), (3) are automatically fulfilled. Hence the system is equivalent to $a+d=0$, $a^2+bc=0$. Given $a$, the numbers $d$ and $bc$ are fixed.