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Proof step 8 from proof ubifs-log-nodups-add4 (previous step)

fo.inodes?, ro.inodes?, dups(log + adr0 ' + adr1 ' + adr2 ' + adr3 '), 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, adr1 = adr2, adr1 = adr3, adr2 = adr3, dups(log), inodekey(0) ∈ ri, adr0 ∈ fs, adr1 ∈ fs, adr2 ∈ fs, adr3 ∈ fs, adr3 ∈ log

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 9:
ro.inodes?, fo.inodes?, dups(log + adr0 ' + adr1 ' + adr2 ' + adr3 '), ro ⊆ dom ri, fo ⊆ dom fi, adr2 ∈ log → adr2 ∈ fs ∧ node-cons(fs[adr2].key, fs[adr2]), ∀ adr. adr ∈ log → adr ∈ fs ∧ node-cons(fs[adr].key, fs[adr]), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ node-closed(fs[ri[key]], ri, fs, ro), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ adr0 = adr3, adr0 = adr2, adr0 = adr1, adr1 = adr3, adr1 = adr2, adr2 = adr3, dups(log), adr3 ∈ log, adr3 ∈ fs, adr2 ∈ fs, adr1 ∈ fs, adr0 ∈ fs, inodekey(0) ∈ ri

Rule argument

pos: left 8 terms: adr2

Additional information

Used terms: adr2