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

adr0 ∈ ri ⊦ ∃ ax1, ax3. adr0 ' + ax + ax0 = ax1 + ax3 ∧ referenced-adrs(ax1, ri) = adr0 ' + referenced-adrs(ax, ri) ∧ referenced-adrs(ax3, ri) = referenced-adrs(ax0, ri)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: adr0 ' + ax, ax0

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]% ⊦ adr ' + ax = adr ' + ax0 ↔ ax = ax0; (s) Used terms: adr0 ' + ax, ax0