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Proof step 10 from proof max-in-ls (previous step)

(m < bag0.max ⊃ bag0.max;m) < n, n ∈ bag0 ++ m, ∀ n0. ¬ bag0.max < n0 ∨ ¬ n0 ∈ bag0 ⊦ bag0 = ∅, m < bag0.max

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m < n, bag0.max < n, n ∈ bag0 ++ m, bag0.max ≤ m, ∀ n0. ¬ bag0.max < n0 ∨ ¬ n0 ∈ bag0 ⊦ bag0 = ∅

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)