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Proof step 6 from proof referenced-adrs-split-one (previous step)

referenced-adrs(ax0, ri) = [], adr0 ∈ ri, ∀ ri, adr. referenced-adrs(ax0, ri) = adr ' → (∃ ax2, ax1. ax0 = ax2 + referenced-adrs(ax0, ri) + ax1 ∧ referenced-adrs(ax2, ri) = [] ∧ referenced-adrs(ax1, ri) = [] ∧ adr ∈ ri) ⊦ ∃ ax, ax1. adr0 ' + ax0 = ax + adr0 ' + ax1 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax1, ri) = []

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], ax0

Additional information

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