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Proof step 19 from proof bag-rem-notmax (previous step)

bag0.max < (bag0 -- m).max, m0 < (bag0 -- m).max, bag0.max ≤ m0, ∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (bag0 -- m).max = m0, bag0 = ∅, bag0 = ∅ ++ m, m = m0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: max-subseteq seq: ⊦ bag0 ≠ ∅ ∧ bag0 ⊆ bag1 → ¬ bag1.max < bag0.max subst: [bag0, bag1] → [bag0 -- m, bag0] flags: <all paths>

Additional information

Used simplifier rules: bag-rem-empty: ⊦ bag ≠ ∅ → (bag -- n = ∅ ↔ bag = ∅ ++ n);%bag-rem-subseteq: ⊦ bag -- n ⊆ bag; Used terms: bag0 -- m, bag0