Entailment relations that are not partial orders The derivability relation of classical logic is a partial order, for which:
\begin{align}
 &(i) &&a \vdash a \\ 
 &(ii) &&\text{If} \hspace{0.2cm} a \vdash b \hspace{0.2cm} \text{and} \hspace{0.2cm} b \vdash a, \hspace{0.2cm} \text{then} \hspace{0.2cm} a = b \\
 &(iii) &&\text{If} \hspace{0.2cm} a \vdash b \hspace{0.2cm} \text{and} \hspace{0.2cm} b \vdash c, \hspace{0.2cm} \text{then} \hspace{0.2cm} a \vdash c  \\ 
\end{align}
Are there logics for which the derivability relation is a pre-order (that is, for which antisymmetry (condition (ii)) does not hold? 
How is equality defined in such logics?
What are such logics like? 
If the substitutivity of provable equivalents in standard first order logic is abandoned, what logic do we get?
The semantic relation of logical consequence of classical logic is a partial order, for which, in any model $\mathscr{M}$ and relative to all variable assignments $g$, we have:
\begin{align}
 &(i') &&a \models_{\mathscr{M, \thinspace g}} \thinspace a \\ 
 &(ii') &&\text{If} \hspace{0.2cm} a \models_{\mathscr{M, \thinspace g}} \thinspace b \hspace{0.2cm} \text{and} \hspace{0.2cm} b \models_{\mathscr{M, \thinspace g}} \thinspace a, \hspace{0.2cm} \text{then} \hspace{0.2cm} a = b \\
 &(iii') &&\text{If} \hspace{0.2cm} a \models_{\mathscr{M, \thinspace g}} \thinspace b \hspace{0.2cm} \text{and} \hspace{0.2cm} b \models_{\mathscr{M, \thinspace g}}
 c, \hspace{0.2cm} \text{then} \hspace{0.2cm} a \models c  \\ 
\end{align}
Are there logics for which the relation of logical consequence is a pre-order (that is, for which antisymmetry (condition (ii')) does not hold? 
How is equality defined in such logics?
 A: Suppose that (in Polish notation) we consider the following axiom sets under uniform substitution and detachment.


*

*{CpCqp, CCpCqrCCpqCpr, CCCpqpp, CCNpqCCNpNqp}.

*{CpCqp, CCpCqrCCpqCpr, CCNpqCCNpNqp}.


Both 1. $\vdash$ 2. and 2. $\vdash$ 1. can get established.  But, axiom set 1. is not independent (CCCpqpp can get proved from the other three), while axiom set 2. is independent. 
So, if we consider sets of axioms in propositional calculi as 'a' and 'b' in (ii), and interpret 'a = b' to mean that the axiom sets both have the same theorems and both have the property of independence, then (ii) can fail.  But, (i) and (iii) hold.  Thus, such logics do seem possible.
A passage in J. Lukasiewicz's Elements of Mathematical Logic establishes that 
CpCpCrCsr $\vdash$ CpCqCrCsr and CpCqCrCsr $\vdash$ CpCpCrCsr.  
Though, they are truth-functionally equal (and Lukasiewicz's notation suggests a truth-functional perspective), Lukasiewicz only calls CpCpCrCsr and CpCqCrCsr inferentially equivalent.  But, as meaningful expressions before a rule of detachment gets introduced, CpCqCrCsr can get used to obtain CpCpCrCsr, but not conversely.  Thus, if derivability gets defined in terms of substitution and detachment, but two forms $\alpha$ and $\beta$ end up equal only on the condition that $\alpha$ can get re-lettered to $\beta$, and $\beta$ can get re-lettered to $\alpha$ also, then (ii) can fail, while (i) and (iii) still hold.
Edit: Consider the following table for some function BE with 0 designated (I got this from Merrie Bergmann's book on Fuzzy Logic who cites a table of Bochvar):
BE  0  1  2
0*  0  2  2  
1   2  0  0     
2   2  0  0

We will say that if BE(x, y) = 0, then x |= y (this works out as valid since if |=BE(x, y) and |=x, then |=y by the first row, the first column, and their intersection).  By the top-left to lower-right diagonal, a |= a.
Now, suppose that a|=b and b|=c.  This breaks down into the following cases:
One. a = 0, b = 0.  By definition a |= b.  0 |= c only holds in the case where c = 0 (BE(0, y) is equal to 0 only when c = 0).  So, if a |= b and b |= c, then a = 0, b = 0, and c = 0.  Since, 0 |= 0, b |= c.  
Two. a = 1, b = 1.  By definition a |= b.  Then, 1 |= c only in the cases where c = 1 and c = 2 by the middle row.  If c = 1, then a |= c is 1 |= 1, which holds.  So, a |= c in this sub-case.  If c = 2, then a |= c is 1 |= 2, which holds.  So, in this sub-case a |= c.  Therefore, if a |= b and b |= c, then a |= c.
Three. a = 1, b = 2.  By definition a |= b.  Then, 2 |= c only in the cases where c = 1 and c = 2 by the bottom row.  Suppose c = 1.  In this sub-case a |= c is 1 |= 1, which is valid.  So, this sub-case holds.  Suppose c = 2.  1 |= 2 holds.  So, this sub-case holds.  Since we covered all sub-cases, therefore, if a |= b and b |= c, then a |= c.
Four. b = 2, a = 2.  By definition a |= b.  Then, 2 |= c only in the cases where c = 1 and c = 2 by the bottom row.  Suppose c = 1.  Then a |= c is 2 |= 1 which holds.  Suppose c = 2.  Then a |= c is 2 |= 2, which holds.  Since we covered both sub-cases, it follows that if a |= b and b |= c, then a |= c.
Since the above four cases suffice, it follows that if a |= b and b |= c, then a |= c for all assignments to a, b, and c.
Thus, (i') and (iii') hold for the above.  But, 1 |= 2 and 2 |= 1, while it is not the case that 2 = 1.  So, (ii') fails for the above.
Equality thus could defined by set-theoretic equality (see my comment to user DanielV) or just the equality of numbers.  Such a logic could have a three-valued model for logical equivalence as Bochvar's logic does.  Though, I will note that the tautologies of the full system of Bochvar's logic that I've referenced here by Bergmann's book has the same set of tautologies as classical logic.   
