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Proof step 20 from proof newref-same (previous step)

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))

User applied rule cut formula

And obtained the following new goals

Proof step 17:
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))
Proof step 19:
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 ⊦ r ∈ st0, next(maxref(st)) = next(maxref(st0))

Rule argument

[fma-arg: r ∈ st0]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (r ∈ st0)]]