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Proof step 3 from proof bag-add2-eq-elim (previous step)

⊦ ¬(∀ a0. count(a0, bag0 ++ a ++ b) = count(a0, bag1 ++ a)) ↔ bag0 ++ b ≠ bag1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a0. (a0 = b ⊃ 1;0) + count(a0, bag0) = count(a0, bag1) ⊦ bag0 ++ b = bag1

Rule argument

Additional information

Used simplifier rules: count-insert: ⊦ count(a, bag ++ b) = (a = b ⊃ 1;0) + count(a, bag);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2