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Proof step 11 from proof newref-same (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
maxref(st0) < maxref(st), r ∈ st0, r ∈ st, maxref(st) ∈ st, maxref(st0) ∈ st0, ∀ r. r ∈ st0 → ¬ maxref(st0) < r, ∀ r. r ∈ st ↔ r ∈ st0, ∀ r. r ∈ st → ¬ maxref(st) < r ⊦ maxref(st) = maxref(st0), maxref(st) ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r < r0 → (¬ r0 < r ↔ true)], [CSIMPSEQ : ⊦ ¬ r < r], [CSIMPFORWARD : ⊦ r < r0 ∧ r0 < r1 → r < r1], [CSIMPSEQ : ⊦ r ≠ r0 → (¬ r < r0 ↔ r0 < r)])]]