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Proof step 5 from proof del-notin (previous step)

count(a, bag0 ++ a0) = 0, ∀ a. count(a, bag0) = 0 → bag0 -- a = bag0 ⊦ (a0 = a ⊃ bag0;(bag0 -- a) ++ a0) = bag0 ++ a0

User applied rule case distinction

And obtained the following new goals

Proof step 6:
count(a, bag0 ++ a0) = 0, ∀ a. count(a, bag0) = 0 → bag0 -- a = bag0 ⊦ (bag0 -- a) ++ a0 = bag0 ++ a0, a0 = a

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ count(a, bag ++ a) = count(a, bag) + 1; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic