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Proof step 1 from proof eq-shift-del (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ (a0 ∈ st[a, d] → st[a, d][a0] = st0[a0])) ↔ (∀ a0. (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0])) ∧ a ∈ st0 ∧ st0[a] = d, a ∈ st

Rule argument

Additional information

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