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Proof step 3 from proof referenced-adrs-split (previous step)

referenced-adrs([], ri) = ax0 + ax1 ⊦ ∃ ax2, ax3. [] = ax2 + ax3 ∧ referenced-adrs(ax2, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ ∃ ax2, ax3. [] = ax2 + ax3 ∧ referenced-adrs(ax2, ri) = [] ∧ referenced-adrs(ax3, ri) = []

Rule argument

Additional information

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