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Proof step 3 from proof Del-at (previous step)

⊦ (st -- b)[a] = st[a], a = b

User applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0[a0, d] -- b)[a] = st0[a0, d][a], a = b

Rule argument

indvar: st

Additional information

Used simplifier rules: gen: ⊦ ;%Del-empty: ⊦ ∅ -- a = ∅;