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Proof step 3 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. (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0])) ∧ a ∈ st0 ∧ st0[a] = d, a ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ (a0 ∈ st[a, d] → st[a, d][a0] = st0[a0]) ⊦ a ∈ st, (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0]) ∧ a ∈ st0 ∧ st0[a] = d

Proof step 3 from proof eq-shift-del (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information