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Proof step 16 from proof newref-same (previous step)

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 ⊦ next(maxref(st)) = next(maxref(st0))

User applied rule cut formula

And obtained the following new goals

Proof step 15:
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), next(maxref(st)) = next(maxref(st0))

Rule argument

[fma-arg: maxref(st) = maxref(st0)]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (maxref(st) = maxref(st0))]]