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Proof step 11 from proof max-subseteq (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: bag0

Additional information

Used terms: bag0