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Proof step 9 from proof lookup-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a0 ∈ ax0, ∀ a, ay. a ∈ ax0 → (ax0 + ay)[a] = ax0[a] ⊦ a0 = a, (ax0 + ay)[a0] = ax0[a0]

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;