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Proof step 6 from proof lookup-append (previous step)

a ∈ p ', ∀ a, ay. a ∈ ax0 → (ax0 + ay)[a] = ax0[a] ⊦ ((p ' + ax0) + ay)[a] = (p ' + ax0)[a]

User applied rule structural induction

And closed the goal!

Rule argument

indvar: p

Additional information

Used simplifier rules: inone: ⊦ a ∈ (b × d0)' ↔ a = b;%assoc-yes: ⊦ ((a × d)' + ax)[a] = d;%pair freely generated by ×;;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;