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Proof step 8 from proof max-in-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m < n, bag0.max < n, m < bag0.max, m + 1 < n, n ∈ bag0 ++ m, ∀ n0. ¬ bag0.max < n0 ∨ ¬ n0 ∈ bag0 ⊦ bag0 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic