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Proof step 8 from proof count-del (previous step)

count(a0, bag0 -- b) = count(a0, bag0) - (a0 = b ∧ count(b, bag0) ≠ 0 ⊃ 1;0) ⊦ count(a0, (a0 = b ⊃ bag0;(bag0 -- b) ++ a0)) = count(a0, bag0) + 1 - (a0 = b ⊃ 1;0)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: sub-base: ⊦ m - 0 = m; from specification nat%count-same: ⊦ count(a, bag ++ a) = count(a, bag) + 1; from specification bag-basic%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2