# Converting a primal-dual pair into an equivalent feasibility problem to prove strong duality in LP

Suppose we have the following primal-dual pair

Primal $$\text{minimize: } c'x$$ Subject to, $$Ax = b$$ $$x\geq0$$

Dual $$\text{maximize: } p'b$$ Subject to, $$p'A \leq c'$$

To convert it into a feasibility problem we do the following:

Converted problem $$\text{maximize: } 0$$ Subject to, $$Ax = b$$ $$p'A \leq c'$$ $$c'x \leq p'b$$ $$x\geq0$$

If there is an optimal solution to the primal-dual problem then the converted problem must be feasible and strong duality holds.

My question is why do we use the constraint $$c'x \leq p'b$$ instead of $$c'x = p'b$$? We know that if this converted problem is feasible then by strong duality the orginal primal-dual pair must have the same cost. How did we guarantee that in the converted problem the feasible solution will satisfy the constraint $$c'x \leq p'b$$ strictly?