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Proof step 6 from proof max-in-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information