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Proof step 10 from proof datanode-pagesize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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