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Proof step 8 from proof max-subseteq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m < bag.max, bag0.max < bag.max, m < bag0.max, m + 1 < bag.max, 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%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m < n → ¬ m + m0 < n; (s)% ⊦ m < n → ¬ n < m; (s)% ⊦ ¬ n < n; (s)% ⊦ m +1 = m + 1; (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