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Proof step 9 from proof subset-smaller (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: bag-basic inst: name: count-other seq: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag) subst: [b, bag, a] → [a, bag, a0] flags: rewrite

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1%count-same: ⊦ count(a, bag ++ a) = count(a, bag) + 1; from specification bag-basic%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: a, bag, a0