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Proof step 9 from proof eq-shift-del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a0 ∈ st ↔ a0 ∈ st0, ∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ (a0 ∈ st[a, d] → st[a, d][a0] = st0[a0]), a0 ∈ st0 → st[a0] = st0[a0] ⊦ a0 = a, a ∈ st, a ∈ st0 ∧ st0[a] = d

Rule argument

Additional information

Used simplifier rules: in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%At-other: ⊦ a ≠ b → st[b, d][a] = st[a];