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Proof step 3 from proof deldel (previous step)

⊦ ∀ a0. (a0 ∈ st -- a -- a ↔ a0 ∈ st -- a) ∧ (st -- a -- a)[a0] = (st -- a)[a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (a0 ∈ st -- a -- a → a0 ∈ st -- a) ∧ (st -- a -- a)[a0] = (st -- a)[a0]

Rule argument

Additional information

Used simplifier rules: Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%Del-notin: ⊦ ¬ a ∈ st -- a;