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Proof step 4 from proof bag-rem-notmax (previous step)

∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ ((bag0 ++ m0) -- m).max = (bag0 ++ m0).max, m = (bag0 ++ m0).max

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (m0 = m ⊃ bag0;(bag0 -- m) ++ m0).max = (bag0 = ∅ ⊃ m0;(m0 < bag0.max ⊃ bag0.max;m0)), m = (bag0 = ∅ ⊃ m0;(m0 < bag0.max ⊃ bag0.max;m0))

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(bag.max > n ⊃ bag.max;n));% ⊦ (bag0 ++ n) -- n0 = (n = n0 ⊃ bag0;(bag0 -- n0) ++ n); (s)% ⊦ m > n ↔ n < m; (s)