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Proof step 9 from proof bag-rm-add-eq (previous step)

⊦ count(a, bag) = 0, (a0 = a ⊃ 1;0) + (count(a0, bag) - (a0 = a ⊃ 1;0)) = count(a0, bag)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: sub-base: ⊦ m - 0 = m; from specification nat%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%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