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Proof step 18 from proof bag-rem-notmax (previous step)

m0 < (bag0 -- m).max, bag0.max ≤ m0, ∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (m0 < (bag0 -- m).max ⊃ (bag0 -- m).max;m0) = m0, bag0 = ∅, bag0 = ∅ ++ m, m = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
bag0.max < (bag0 -- m).max, m0 < (bag0 -- m).max, bag0.max ≤ m0, ∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (bag0 -- m).max = m0, bag0 = ∅, bag0 = ∅ ++ m, m = m0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic