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Proof step 10 from proof newref-same (previous step)

maxref(st0) < maxref(st), r ∈ st0, r ∈ st, maxref(st) ∈ st, maxref(st0) ∈ st0, ∀ r. r ∈ st0 → ¬ maxref(st0) < r, ∀ r. r ∈ st ↔ r ∈ st0, ∀ r. r ∈ st → ¬ maxref(st) < r ⊦ maxref(st) = maxref(st0), maxref(st) ∈ st0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 7, [QUANTTERMLIST : (maxref(st))]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (maxref(st))]]