Logics in which bi-implication and equivalence comes apart Where $\phi, \psi$ are variables over formulas, are there logics for which we have either (or both) of $(1)$ and $(2)\thinspace$?
$$(1) \hspace{0.5cm}(\phi \rightarrow \psi)\hspace{0.3cm} \land \hspace{0.3cm} (\psi \rightarrow \phi)\hspace{0.3cm} \nvdash \hspace{0.3cm} \phi \hspace{0.3cm} \leftrightarrow \hspace{0.3cm} \psi $$
$$(2) \hspace{0.5cm}(\phi \rightarrow \psi)\hspace{0.3cm} \land \hspace{0.3cm} (\psi \rightarrow \phi)\hspace{0.3cm} \not\models \hspace{0.3cm} \phi \hspace{0.3cm} \leftrightarrow \hspace{0.3cm} \psi $$
Are there logics for which we have either (or both) of $(3)$ and $(4)$?
$$(3) \hspace{0.3cm} \phi \hspace{0.3cm} \leftrightarrow \hspace{0.3cm} \psi \hspace{0.3cm} \nvdash \hspace{0.3cm} \hspace{0.5cm}(\phi \rightarrow \psi)\hspace{0.3cm} \land \hspace{0.3cm} (\psi \rightarrow \phi)  $$
$$(4) \hspace{0.3cm} \phi \hspace{0.3cm} \leftrightarrow \hspace{0.3cm} \psi \hspace{0.3cm} \not\models \hspace{0.3cm} \hspace{0.5cm}(\phi \rightarrow \psi)\hspace{0.3cm} \land \hspace{0.3cm} (\psi \rightarrow \phi)  $$
In such logics the equivalence of two formulas $\thinspace \phi, \thinspace \psi \thinspace$ (represented by $\leftrightarrow$) would come apart from their being derivable from one another.
In which setting might such a logic be desirable?

Edit: Would it be correct to think of such a logic as a logic in which $\rightarrow$ is not a partial order, but rather a pre-order $\textit{with respect to the $\equiv$ relation}$, where $\equiv$ replaces the usual symbol = used in the statement of antisymmetry (see $(iii)$), and where $\equiv$ is a mere notational variant of $\leftrightarrow$ as used above (chosen $\textbf{for clarity}$)? 
That is would it be correct to say that $\rightarrow$ in logics for which (1), (2), (3) or (4) fails is a pre-order in the sense that we have $(i)$ and $(ii)$ but not $(iii)$ (antisymmetry fails)? $\\$
$$(i) \hspace{0.3cm} \vdash \hspace{0.3cm} a ≤ a \hspace{0.3cm} (reflexivity)$$
$$(ii) \hspace{0.3cm} \vdash \hspace{0.3cm} if \hspace{0.3cm} a ≤ b \hspace{0.3cm}and \hspace{0.3cm}  b ≤ c \hspace{0.3cm} then \hspace{0.3cm} a ≤ c  \hspace{0.3cm}(transitivity)$$
$$(iii) \hspace{0.3cm} \nvdash \hspace{0.3cm} If \hspace{0.3cm} a ≤ b \hspace{0.3cm} and \hspace{0.3cm} b ≤ a \hspace{0.3cm} then \hspace{0.3cm} a \equiv b \hspace{0.3cm} (no \hspace{0.3cm} Antisymmetry)$$
 A: There are logic in which there is no conjunction. In that case only only of $\to$ or $\leftrightarrow$ should be there, otherwise conjunction can be defined via $\phi\land\psi:=(\phi \to \psi) \leftrightarrow \psi$.
The study of such logics can be interesting from computational complexity point of view. Deciding tautology for intuitionistic propositional logic can be reduced to deciding tautology for the implicational fragment of intuitionistic (minimal) logic. Deciding tautology for the implicational fragment of intuitionistic (or minimal) propositional logic is PSPACE complete.

Regarding the question about semantics definitions for $\to$ and $\leftrightarrow$ by aphorisme and Mees de Vries, one example of a large class of propositional logics can be derived from an idempotent and increasing operator $\operatorname{op}$:


*

*$\operatorname{op}(\operatorname{op}(X))=\operatorname{op}(X)$

*$X \subseteq Y \Rightarrow \operatorname{op}(X) \subseteq \operatorname{op}(Y)$


The propositions would be the subsets $X$ with $X=\operatorname{op}(X)$ and operations would be defined by translating logical operations into operation on sets (pointwise...), and applying $\operatorname{op}$ to the result.
For the question at hand, this would translate into whether
$$\operatorname{op}(\operatorname{op}(A^c\cup B)\cap\operatorname{op}(A\cup B^c))=\operatorname{op}((A^c\cup B)\cap(A\cup B^c))$$
This is probably closely related to 


*

*$\operatorname{op}(X)\subseteq X$


i.e. it will in general only be true intensive operators $\operatorname{op}$. (Exercise: Can you show that for intensive increasing operators $\operatorname{op}$ we have $\operatorname{op}(A\cap B)=\operatorname{op}(\operatorname{op}(A)\cap\operatorname{op}(B))$? What about the converse?)
