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Proof step 5 from proof Del-at (previous step)

a = a0, ∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0[a0, d] -- b)[a] = st0[a0, d][a], a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0[a0, d] -- b)[a0] = d, a0 = b

Rule argument

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d;