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Proof step 7 from proof dentry-ubifs-cons (previous step)

fo.inodes?, ro.inodes?, log-cons(log, fs), ri-closed(ri, fs, ro), ri[dentrykey(ino, str)] ∈ fs, dentrykey(ino, str) ∈ ri, inodekey(ino) ∈ ri, fo ⊆ dom fi, ro ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ fs[ri[dentrykey(ino, str)]].dentry?, dups(log), inodekey(0) ∈ ri

User applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: dentrykey(ino, str) abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ key-cons(dentrykey(ino, str), nd) ↔ nd.dentry? ∧ nd.key = dentrykey(ino, str); (s) Used terms: dentrykey(ino, str)