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Proof step 3 from proof insert-01 (previous step)

a ∈ s ⊦ # s = 0, st[b, d][a] = st[a], s = ∅, b ∈ s

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
a ∈ s0 ++ a0, ∀ a, st, b, d. a ∈ s0 → # s0 = 0 ∨ st[b, d][a] = st[a] ∨ s0 = ∅ ∨ b ∈ s0 ⊦ #(s0 ++ a0) = 0, s0 ++ a0 = ∅, st[b, d][a] = st[a], b ∈ s0 ++ a0

Rule argument

term: s

Additional information

Used simplifier rules: set generated by ∅, ++;;