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Proof step 6 from proof inode-ubifs-cons (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: inodekey(ino) abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ key-cons(inodekey(ino), nd) ↔ nd.inode? ∧ nd.key = inodekey(ino) ∧ ino ≠ 0; (s) Used terms: inodekey(ino)