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Proof step 5 from proof referenced-adrs-nodups (previous step)

dups((adr ∈ ri ⊃ adr ';[]) + referenced-adrs(ax0, ri)), ∀ ri0. ¬ dups(referenced-adrs(ax0, ri0)) ⊦ dups(ax0), adr ∈ ax0

User applied rule all left

And obtained the following new goals

Proof step 6:
dups((adr ∈ ri ⊃ adr ';[]) + referenced-adrs(ax0, ri)) ⊦ dups(ax0), dups(referenced-adrs(ax0, ri)), adr ∈ ax0

Rule argument

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

Additional information

Used terms: ri