A partial operation on a nonempty set $A$ is a map $f:\mathrm{dom}(f,A)\to A$ where $\mathrm{dom}(f,A)\subseteq A^n$ for some $n\in\mathbb{N}$. A partial algebra is an ordered pair $(A,P)$ where $A$ is a nonempty set and $P$ is a set of partial operations on $A$. A congruence on $\mathfrak{A}=(A,P)$ is an equivalence relation $\sim$ such that if $f\in P$, $\mathrm{dom}(f,A)\subseteq A^n$, $(a_1,\ldots, a_n)$,$(b_1,\ldots,b_n)\in\mathrm{dom}(f,A)$, and $a_1\sim b_1, \ldots, a_n\sim b_n$, then $f(a_1,\ldots,a_n)\sim f(b_1,\ldots, b_n)$. Let $\mathrm{Con}\mathfrak{A}$ denote the set of all congruences on $\mathfrak{A}$. I want to prove that $\mathrm{Con}\mathfrak{A}$ is an algebraic lattice.
The problem is that the join of two partial algebra congruences (as equivalence relations) is not necessarily a congruence. Preferably, I would like a proof that avoids the use of directed unions or closure operators.