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Proof step 4 from proof subset-empty-size (previous step)

bag = bag0 ++ a, bag ⊆ ∅ ⊦ # bag = 0, bag = bag0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
bag0 ++ a ⊆ ∅ ⊦ bag0 ++ a = bag0

Rule argument

Additional information

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