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Proof step 2 from proof same-subclasses-classex-02 (previous step)

class∃(str0, class), class ≤ class0, ∀ class1. interface∃(str0, class1) ∧ class1 ≤ class0 → interface∃(str, class1), ∀ class1. class∃(str0, class1) ∧ class1 ≤ class0 → class∃(str, class1) ⊦ class∃(str, class)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 4, [QUANTTERMLIST : (class)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (class)]]