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Proof step 5 from proof lookup-append (
previous step
)
a ∈ p ', ∀ a, ay. a ∈ ax0 → (ax0 + ay)[a] = ax0[a] ⊦ ((p ' + ax0) + ay)[a] = (p ' + ax0)[a], ¬ a ∈ p ' ∧ ¬ a ∈ ax0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
a ∈ p ', ∀ a, ay. a ∈ ax0 → (ax0 + ay)[a] = ax0[a] ⊦ ((p ' + ax0) + ay)[a] = (p ' + ax0)[a]
Rule argument
Additional information