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Proof step 4 from proof referenced-adrs-split (previous step)

⊦ ∃ ax2, ax3. [] = ax2 + ax3 ∧ referenced-adrs(ax2, ri) = [] ∧ referenced-adrs(ax3, ri) = []

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], []

Additional information

Used simplifier rules: referenced-adrs-base: ⊦ referenced-adrs([], ri) = [];% ⊦ [] + ax = ax; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set] Used terms: [], []