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Proof step 4 from proof rem-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ st, ∀ a0. a0 ∈ st -- a ↔ a0 ∈ st ⊦

Rule argument

Additional information

Used simplifier rules: notin-del-notin: ⊦ ¬ a ∈ st → st -- a = st;%Del-notin: ⊦ ¬ a ∈ st -- a;%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a];