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Proof step 11 from proof eq-shift-del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a ∈ st0, ∀ a0. (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0]) ⊦ a ∈ st, (a0 ∈ st[a, st0[a]] ↔ a0 ∈ st0) ∧ (a0 ∈ st[a, st0[a]] → st[a, st0[a]][a0] = st0[a0])

Proof step 11 from proof eq-shift-del (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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