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Proof step 7 from proof eq-shift-del (previous step)

∀ 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 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0]) ∧ a ∈ st0 ∧ st0[a] = d

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ 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

Rule argument

Additional information

Used simplifier rules: Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%Del-at: ⊦ a ≠ b → (st -- b)[a] = st[a];