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Proof step 3 from proof at-diffs (previous step)

r ∈ H ⊦ (H \ rs)[r] = H[r], r ∈ rs

User applied rule structural induction

And obtained the following new goals

Proof step 4:
r ∈ H, ∀ r, H. r ∈ H → (H \ rs0)[r] = H[r] ∨ r ∈ rs0 ⊦ (H \ (rs0 ++ r0))[r] = H[r], r ∈ rs0 ++ r0

Rule argument

indvar: rs

Additional information

Used simplifier rules: diffs-emp: ⊦ H0 \ ∅ = H0;%refset generated by ∅, ++;;