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Proof step 5 from proof lookup-append-01 (previous step)

⊦ (∀ ay, a. (ax0 + ay)[a] = ay[a] ∨ a ∈ ax0) → (((a × d)' + ax0) + ay)[a0] = ay[a0] ∨ a0 ∈ (a × d)' ∨ a0 ∈ ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ ay, a. (ax0 + ay)[a] = ay[a] ∨ a ∈ ax0 ⊦ a0 = a, (ax0 + ay)[a0] = ay[a0], a0 ∈ ax0

Rule argument

Additional information

Used simplifier rules: inone: ⊦ a ∈ (b × d0)' ↔ a = b;%assoc-no: ⊦ a ≠ b → ((b × d)' + ax)[a] = ax[a];% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;