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Proof step 7 from proof max-subseteq (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
bag0.max < (bag = ∅ ⊃ m;(m < bag.max ⊃ bag.max;m)), m < bag0.max, bag ⊆ bag0, (∃ bag1. bag1.max < bag.max ∧ bag ⊆ bag1) → bag = ∅ ⊦ bag0 = ∅
Proof step 10:
m < (bag = ∅ ⊃ m;(m < bag.max ⊃ bag.max;m)), bag ⊆ bag0, (∃ bag1. bag1.max < bag.max ∧ bag ⊆ bag1) → bag = ∅ ⊦ bag0 = ∅, m < bag0.max

Rule argument

pos: left 1

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