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Proof step 5 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 ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
(bag0 = ∅ ⊃ m;(m < bag0.max ⊃ bag0.max;m)) < n, n ∈ bag0 ++ m, ∀ n0. ¬ bag0.max < n0 ∨ ¬ n0 ∈ bag0 ⊦ bag0 = ∅

Rule argument

fma: bag0 = ∅

Additional information

Used simplifier rules: bag-in-single: ⊦ n ∈ ∅ ++ m ↔ n = m;% ⊦ ¬ n ∈ ∅; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s) Used formulas: bag0 = ∅