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Proof step 11 from proof datanode-pagesize (previous step)

fo.inodes?, ro.inodes?, datakey(m, n) ∈ ri, inodekey(m) ∈ ri, ri[datakey(m, n)] ∈ fs, fo ⊆ dom fi, ro ⊆ dom ri, fs[ri[datakey(m, n)]].dentry? → inodekey(fs[ri[datakey(m, n)]].key.ino) ∈ ri ∧ ¬ inodekey(fs[ri[datakey(m, n)]].key.ino) ∈ ro ∧ fs[ri[inodekey(fs[ri[datakey(m, n)]].key.ino)]].directory ∧ inodekey(fs[ri[datakey(m, n)]].ino) ∈ ri ∧ ¬ inodekey(fs[ri[datakey(m, n)]].ino) ∈ ro, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∀ adr. adr ∈ log → adr ∈ fs ∧ node-cons(fs[adr].key, fs[adr]) ⊦ # fs[ri[datakey(m, n)]].data = VFS_PAGE_SIZE, fs[ri[datakey(m, n)]].data?, fs[ri[inodekey(m)]].directory, dups(log), inodekey(0) ∈ ri

User applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: datakey(m, n) abort: F discard: T computed: T

Additional information

Used simplifier rules: datakey-cons: ⊦ key-cons(datakey(ino, n), nd) ↔ nd.data? ∧ nd.key = datakey(ino, n); Used terms: datakey(m, n)