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Proof step 20 from proof ubifs-log-nodups-add4 (previous step)

fo.inodes?, ro.inodes?, dups(log + adr0 ' + adr1 '), adr1 ∈ log, fo ⊆ dom fi, ro ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ node-closed(fs[ri[key]], ri, fs, ro), ∀ adr. adr ∈ log → adr ∈ fs ∧ node-cons(fs[adr].key, fs[adr]) ⊦ adr0 = adr1, adr0 = adr2, adr0 = adr3, adr2 = adr3, dups(log), inodekey(0) ∈ ri, adr0 ∈ fs, adr1 ∈ fs, adr2 ∈ fs, adr3 ∈ fs, adr3 ∈ log, adr2 ∈ log

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: adr1

Additional information

Used terms: adr1