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Proof step 7 from proof count-del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%count-same: ⊦ count(a, bag ++ a) = count(a, bag) + 1; from specification bag-basic