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Proof step 2 from proof referenced-adrs-eq-one-lookup (previous step)

referenced-adrs(ax, ri) = adr ' ⊦ fs[referenced-adrs(ax, ri)] = fs[adr] '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
referenced-adrs(ax1, ri) = [], referenced-adrs(ax0, ri) = [], adr ∈ ri ⊦ ∃ adr0. (∃ ax2, ax3. ax0 + adr ' + ax1 = ax2 + adr0 ' + ax3 ∧ referenced-adrs(ax2, ri) = [] ∧ referenced-adrs(ax3, ri) = [] ∧ adr0 ∈ ri) ∧ fs[adr0] = fs[adr]

Rule argument

Additional information

Used simplifier rules: referenced-adrs-split-one: ⊦ referenced-adrs(ax, ri) = adr ' ↔ (∃ ax0, ax1. ax = ax0 + adr ' + ax1 ∧ referenced-adrs(ax0, ri) = [] ∧ referenced-adrs(ax1, ri) = [] ∧ adr ∈ ri);% ⊦ fs[ax] = nd ' ↔ (∃ adr. ax = adr ' ∧ fs[adr] = nd); (s)