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Proof step 6 from proof in-init-02 (previous step)

r -- sk ∈ addarray(r0, vals0, j + 1, st), r ∈ st, ∀ r, sk, r0, j, st. r -- sk ∈ addarray(r0, vals0, j, st) ∧ r ∈ st → r = r0 ∨ r -- sk ∈ st ⊦ r = r0, r -- sk ∈ st

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 3, [QUANTTERMLIST : (r, sk, r0, j + 1, st)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, sk, r0, j + 1, st)]]