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Proof step 4 from proof eq-shift-put (previous step)

a ∈ st ⊦ (∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ st[a, d][a0] = st0[a0]) ↔ st = st0[a, st[a]] ∧ a ∈ st0 ∧ st0[a] = d

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information