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Proof step 9 from proof newref-same (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 8:
maxref(st) < maxref(st0), r ∈ st0, r ∈ st, maxref(st) ∈ st, maxref(st0) ∈ st0, ∀ r. r ∈ st ↔ r ∈ st0, maxref(st0) ∈ st → ¬ maxref(st) < maxref(st0), ∀ r. r ∈ st → ¬ maxref(st) < r, ∀ r. r ∈ st0 → ¬ maxref(st0) < r ⊦ maxref(st) = maxref(st0)

Rule argument

[EXRARG : pos: left 7, [extquanttermlist: (maxref(st0)) abort:F , discard:F , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (maxref(st0))]]