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Proof step 6 from proof exin (previous step)

⊦ count(a, bag) = 0, count(a0, bag) = count(a0, (bag -- a) ++ a), ∃ bag0. bag = bag0 ++ a

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 -- a, a0] flags: rewrite

Additional information

Used simplifier rules: count-del: ⊦ count(a, bag -- b) = count(a, bag) - (a = b ∧ count(a, bag) ≠ 0 ⊃ 1;0);%count-del: ⊦ count(a, bag -- b) = count(a, bag) - (a = b ∧ count(a, bag) ≠ 0 ⊃ 1;0);%sub-base: ⊦ m - 0 = m; from specification nat%count-same: ⊦ count(a, bag ++ a) = count(a, bag) + 1; from specification bag-basic%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2 Used terms: a, bag -- a, a0