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Proof step 4 from proof Del-in (previous step)

b = a0, ∀ a, b. a ∈ st0 -- b ↔ a ≠ b ∧ a ∈ st0 ⊦ a ∈ st0[a0, d] -- b ↔ a ≠ b ∧ a ∈ st0[a0, d]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a, b. a ∈ st0 -- b ↔ a ≠ b ∧ a ∈ st0 ⊦ a ∈ st0 -- a0 ↔ a ≠ a0 ∧ a ∈ st0

Rule argument

Additional information

Used simplifier rules: Del-rec-same: ⊦ st[a, d] -- a = st -- a;%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);