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Proof step 4 from proof max-in-ls (previous step)

(bag0 ++ m).max < n, n ∈ bag0 ++ m, ∀ n. ¬(bag0.max < n ∧ n ∈ bag0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(bag0 = ∅ ⊃ m;(m < bag0.max ⊃ bag0.max;m)) < n, n ∈ bag0 ++ m, ∀ n0. ¬ bag0.max < n0 ∨ ¬ n0 ∈ bag0 ⊦

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(bag.max > n ⊃ bag.max;n));% ⊦ m > n ↔ n < m; (s)