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Proof step 19 from proof subset-plusminus (previous step)

count(a0, bag1) < count(a0, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, count(a0, bag0) = 0, a0 = a

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a0 abort: F discard: F computed: F

Additional information

Used simplifier rules: count-other: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag); from specification bag-basic%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: a0