Turning a bra made up of a tensor product of two bras into a ket (and vice-versa)

I know that in general the following statement is true: $$langlephi|chirangle = langlechi|phirangle^* $$

And for the operator $A$ then the following identity also holds:
$$ langle psi| A|phirangle = langlephi|A^dagger |psi rangle$$

Does this mean that (1) implies (2)

$$| psirangle = |chirangle|phirangletag1$$
$$langle psi|= langlephi|langlechi|tag2$$

since I assume$$ langlepsi|psirangle = 1?$$