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Proof step 13 from proof Del-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0 -- b)[a] = st0[a], a = a0, a = b, b = a0

Rule argument

Additional information

Used simplifier rules: At-other: ⊦ a ≠ b → st[b, d][a] = st[a];