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Proof step 2 from proof inex (previous step)

a ∈ st ⊦ ∃ d, st0. st = st0[a, d] ∧ ¬ a ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ st ⊦ ∃ d, st0. (∀ a0. (a0 ∈ st ↔ a0 ∈ st0[a, d]) ∧ st[a0] = st0[a, d][a0]) ∧ ¬ a ∈ st0

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]);