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Proof step 10 from proof max-subseteq (previous step)

m < (bag = ∅ ⊃ m;(m < bag.max ⊃ bag.max;m)), bag ⊆ bag0, (∃ bag1. bag1.max < bag.max ∧ bag ⊆ bag1) → bag = ∅ ⊦ bag0 = ∅, m < bag0.max

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m < (m < bag.max ⊃ bag.max;m), bag0.max < bag.max, bag0.max ≤ m, bag ⊆ bag0 ⊦ bag = ∅, bag0 = ∅, ∃ bag1. bag1.max < bag.max ∧ bag ⊆ bag1

Rule argument

Additional information

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