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Proof step 5 from proof max-subseteq (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ bag0 ⊆ ∅ ↔ bag0 = ∅; (s)% ⊦ ¬ n < n; (s)