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Proof step 18 from proof referenced-adrs-split (previous step)

⊦ adr ∈ ri, ∃ ax1, ax4. adr ' + ax + ax3 = ax1 + ax4 ∧ referenced-adrs(ax1, ri) = referenced-adrs(ax, ri) ∧ referenced-adrs(ax4, ri) = referenced-adrs(ax3, ri)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: adr ' + ax, ax3

Additional information

Used simplifier rules: referenced-adrs-rec: ⊦ referenced-adrs(adr ' + ax, ri) = (adr ∈ ri ⊃ adr ';[]) + referenced-adrs(ax, ri);%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ [] + ax = ax; (s) Used terms: adr ' + ax, ax3