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Proof step 359 from proof inv-truncate (previous step)

m = 0 ∧ ino ≠ ROOT_INO ↔ inodekey(ino) ∈ ro, VFS_PAGE_SIZE = n0 + m4 % VFS_PAGE_SIZE, fs0[ri0[datakey(ino, m4 / VFS_PAGE_SIZE)]] = datanode(datakey(ino, m4 / VFS_PAGE_SIZE), buf0), fs0[ri0[inodekey(ino)]] = inodenode(key1, md, false, m, m1, m2), ro.inodes?, fo.inodes?, m4 / VFS_PAGE_SIZE * VFS_PAGE_SIZE < m4, m4 / VFS_PAGE_SIZE * VFS_PAGE_SIZE < n1, file-node-cons(fs0, ri0), log-cons(log0, fs0), orphans-cons(fs0, ri0, ro), replay-cons-strong(fs0, ri0, ro, fi, fo, log0), ri-closed(ri0, fs0, ro), ri-cons(ri0, fs0), ri-inj(ri0), ri-inj(fi), ri0[inodekey(ino)] ∈ fs0, datakey(ino, m4 / VFS_PAGE_SIZE) ∈ ri0, ri0[datakey(ino, m4 / VFS_PAGE_SIZE)] ∈ fs0, inodekey(ino) ∈ ri0, m4 ≤ n1, ro ⊆ dom ri0, fo ⊆ dom fi ⊦ # buf0 = VFS_PAGE_SIZE, m4 % VFS_PAGE_SIZE = 0, adr1 = adr2, adr1 = adr3, adr2 = adr3, ino = 0, buf-opt.some?, dups(log0), adr1 ∈ fs0, adr2 ∈ fs0, adr3 ∈ fs0, adr3 ∈ ri0, inodekey(0) ∈ ri0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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