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Proof step 17 from proof newref-same (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
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))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]