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Proof step 17 from proof eq-shift-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
a ∈ st0, a ∈ st, ∀ a0. (a0 ∈ st ↔ a0 ∈ st0[a, st[a]]) ∧ st[a0] = st0[a, st[a]][a0] ⊦ (a0 ∈ st[a, st0[a]] ↔ a0 ∈ st0) ∧ st[a, st0[a]][a0] = st0[a0]

Proof step 17 from proof eq-shift-put (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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