p(x) ∧ q(x) → t() and only one piece of evidence p(C) there are no relevant groundings because there is no evidence for q(C), and therefore, for normal PSL, I(p(C) ∧ q(C)) = 0 which does not affect I(t()).
when using averaging with the same evidence, we need to generate the grounding p(C)∧q(C) because I(p(C)∧q(C)) = 0.5 which does affect I(t()).