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Proof step 3 from proof count-del (previous step)

∀ a, b. count(a, bag0 -- b) = count(a, bag0) - (a = b ∧ count(b, bag0) ≠ 0 ⊃ 1;0) ⊦ count(a, (bag0 ++ a0) -- b) = 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 4:
∀ a, b. 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)

Rule argument

Additional information

Used simplifier rules: del-rec: ⊦ (bag ++ a) -- b = (a = b ⊃ bag;(bag -- b) ++ a);