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Proof step 4 from proof insert-02 (previous step)

a ∈ s0, ∀ a1, st0, d0, b0. a1 ∈ s0 → st0[a1, d0][b0] = st0[b0] ∨ b0 ∈ s0 ⊦ # s0 = 0, b = a0, #(s0 ++ a0) = 0, st[a, d][b] = st[b], a = a0, s0 = ∅, b ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a, st, d, b

Additional information

Used terms: a, st, d, b