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Proof step 2 from proof Diff-get (previous step)

⊦ (st \ s)[a] = st[a], a ∈ s

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ st, a. (st \ s0)[a] = st[a] ∨ a ∈ s0 ⊦ (st \ (s0 ++ a0))[a] = st[a], a ∈ s0 ++ a0

Rule argument

indvar: s

Additional information

Used simplifier rules: Diff-base: ⊦ st \ ∅ = st;%set generated by ∅, ++;;