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Proof step 20 from proof referenced-adrs-split-one (previous step)

referenced-adrs(ax0, ri) = adr ', ∀ 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) ⊦ adr ∈ ri, adr0 ∈ ri

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: ri, adr abort: F discard: T computed: T

Additional information

Used terms: ri, adr