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Proof step 14 from proof referenced-adrs-split-one (previous step)

referenced-adrs(ax3, ri) = [], adr1 ∈ ri, ∀ ri0, adr2. (adr1 ∈ ri0 ⊃ adr1 ';[]) + referenced-adrs(ax3, ri0) = adr2 ' → (∃ ax, ax0. adr1 ' + ax3 = ax + ((adr1 ∈ ri0 ⊃ adr1 ';[]) + referenced-adrs(ax3, ri0)) + ax0 ∧ referenced-adrs(ax, ri0) = [] ∧ referenced-adrs(ax0, ri0) = [] ∧ adr2 ∈ ri0) ⊦ adr0 ∈ ri, ∃ ax, ax0. adr0 ' + adr1 ' + ax3 = ax + adr1 ' + ax0 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax0, ri) = []

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: adr0 ', ax3

Additional information

Used simplifier rules: referenced-adrs-one: ⊦ referenced-adrs(adr ', ri) = (adr ∈ ri ⊃ adr ';[]);%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set] Used terms: adr0 ', ax3