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Proof step 3 from proof lookup-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a, ay. a ∈ ax0 → (ax0 + ay)[a] = ax0[a] ⊦ ((p ' + ax0) + ay)[a] = (p ' + ax0)[a], ¬ a ∈ p ' ∧ ¬ a ∈ ax0

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p + ax = p ' + ax;