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Proof step 5 from proof same-subclasses-subclass (previous step)

interface∃(str0, class2), class2 ≤ class0, class0 ≤ class, class2 ≤ class, ∀ class0. interface∃(str0, class0) ∧ class0 ≤ class → interface∃(str, class0), ∀ class0. class∃(str0, class0) ∧ class0 ≤ class → class∃(str, class0) ⊦ interface∃(str, class2)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 5, [QUANTTERMLIST : (class2)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (class2)]]