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Proof step 13 from proof newref-same (previous step)

maxref(st) < maxref(st0) ∨ maxref(st0) < maxref(st), 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 case distinction

And obtained the following new goals

Proof step 9:
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)
Proof step 12:
maxref(st0) < maxref(st), 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)

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]