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Proof step 10 from proof count-del (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: count-other: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag); from specification bag-basic%sub-base: ⊦ m - 0 = m; from specification nat