Consider an Integer Linear Program (ILP) of the following form:
ILP 1: Minimize $c^1x^1 + c^2 x^2+ c^3 x^3$
Subject to:
\begin{align} A^1x^1 &\leq b^1\\ A^2x^2 &\leq b^2\\ A^3_1x^1 + A^3_2x^2 + A^3_3x^3 &\leq b^3\\ x^1, x^2, x^3 \geq 0\\ x^1, x^2, x^3 \;\text{are integer vectors} \end{align}
We have the following relaxations.
LP ($p$): Minimize $c^1x^1 + c^2 x^2+ c^3 x^3$
Subject to:
\begin{align} A^1x^1 &\leq b^1\\ A^3_1x^1 + A^3_2x^2 + A^3_3x^3 &\leq b^3\\ x^1, x^2, x^3 \geq 0 \end{align}
Solving LP $p$ gives integral solutions $x^1_p, x^2_p=0, x^3_p$.
LP ($q$): Minimize $ c^2 x^2+ c^3 x^3$
Subject to:
\begin{align} A^2x^2 &\leq b^2\\ A^3_2x^2 + A^3_3x^3 &\leq b^3\\ x^1, x^2, x^3 \geq 0 \end{align}
Solving LP $q$ gives a solution $x^2_q, x^3_q$, not necessarily integral.
LP ($r$): Minimize $c^1x^1 +c^2x^2+ c^3 x^3$
Subject to:
\begin{align} A^1x^1 &\leq b^1\\ A^3_1x^1 + A^3_2(x^2_q+x^2) + A^3_3(x^3_q+x^3) &\leq b^3\\ x^1, x^2, x^3 \geq 0 \end{align}
Solving LP $r$ gives integral solutions $x^1_r=x^1_p, x^2_r=0, x^3_r=x^3_p$.
Then is $c^1x^1_p+c^3x^3_p+c^2x^2_q+c^3x^3_q$ lower bound for the ILP? It seems intuitive to me because of the result in LP $r$, but I am not able to give a formal proof. If it is not a lower bound then is there a simple condition under which this will be true?
Additional Information: The key aspect of the problem is the condition from LP $r$. This property is proved using duality. Of course, if we don't consider the condition, the value of $c^1x^1_p+c^3x^3_p+c^2x^2_q+c^3x^3_q$ is not a lower bound as shown in one of the answers by @prubin.
No, the sum is not a lower bound. Here is a counterexample, in which all variables are scalar: \begin{align*} \min\quad & x_{1}+x_{2}+x_{3}\\ \textrm{s.t.}\quad & x_{1}\le5\\ & x_{2}\le5\\ & x_{1}+x_{2}-2x_{3}\le-6\\ & x\in\mathbb{Z}_{\ge0}^{3}. \end{align*} By inspection, the optimal solution is $(0,0,3)$ with objective value 3. Your first LP is \begin{align*} \min\quad & x_{1}+x_{2}+x_{3}\\ \textrm{s.t.}\quad & x_{1}\le5\\ & x_{1}+x_{2}-2x_{3}\le-6\\ & x\ge 0 \end{align*}with optimal solution $(0,0,3)$ and objective value 3. Your second LP is \begin{align*} \min\quad & x_{2}+x_{3}\\ \textrm{s.t.}\quad & x_{2}\le5\\ & x_{2}-2x_{3}\le-6\\ & x\ge 0 \end{align*}with optimal solution $(x_2,x_3)=(0,3)$ and optimal value 3. Add the objective values together and you get 6, which is not a lower bound for the value (3) of the ILP.
As to conditions under which you would get a valid lower bound, I have no suggestions.