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Proof step 1 from proof referenced-adrs-split (conclusion)

⊦ referenced-adrs(ax, ri) = ax0 + ax1 ↔ (∃ ax2, ax3. ax = 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 2:
referenced-adrs(ax, ri) = ax0 + ax1 ⊦ ∃ ax2, ax3. ax = ax2 + ax3 ∧ referenced-adrs(ax2, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1

Rule argument

Additional information

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