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Proof step 234 from proof eq-io-truncate (previous step)

files1[m] = mkfile(m0, md0, m1, content0), dirs = dirs1, dirs0 = dirs1, files = files1[m, mkfile(m0, md0, m1, content0 upto min(n0, m2))], ino = inode.ino, inode = mkinode(m, md, false, m1, 0, n0), m ∈ files1, buf-opt0.some? → # buf-opt0.buf = VFS_PAGE_SIZE ⊦ ⟨asynced := ?⟩ ( (m ∈ dirs ⊃ mkinode(m, dirs[m].meta, true, dirs[m].nlink, dirs[m].nsubdirs, dirs[m].size) ;mkinode(m, files[m].meta, false, files[m].nlink, 0, files[m].size) ) = (m ∈ dirs0 ⊃ mkinode(m, dirs0[m].meta, true, dirs0[m].nlink, dirs0[m].nsubdirs, dirs0[m].size);mkinode(m, md, false, m1, 0, n0)) ∧ ESUCCESS = ESUCCESS ∧ dirs = dirs0 ∧ files = files1[m, mkfile(n0, md, m1, content0 upto min(n0, m2))]), ino ∈ dirs, ino ∈ files, m2 ≤ n0 ∧ m2 % VFS_PAGE_SIZE ≠ 0 ∧ (m2 / VFS_PAGE_SIZE ∈ content0 ∨ buf-opt0.some?)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkinode(m, md, boolvar, m0, m1, m2).ino = m; (s)% ⊦ ino ∈ files[ino0, af] ↔ ino ∈ files; (s)