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Proof step 175 from proof truncate-inv (previous step)

⟨files := files[ino, files[ino].size:= inode.size][ino, files[ino, files[ino].size:= inode.size][ino].meta:= inode.meta]⟩ (dirs = dirs0 ∧ files = files0), files2[ino4] = mkfile(m0, md, # links(ino4, dirs), content0), files0[ino4] = mkfile(m, md0, m1, content1), files = files1, files1 = files2[ino4, mkfile(m0, md, # links(ino4, dirs), content0 upto m0)], ino = inode.ino, inode = mkinode(ino4, md, false, # links(ino4, dirs), 0, m0), afs-cons(dirs0, files0), afs-cons(dirs, files2), files2 ~ files0, ino ∈ files, ino4 ∈ files2, n0 ≤ # buf0 ⊦ ino4 ∈ files0 ∧ m = m0, m0 % VFS_PAGE_SIZE ≠ 0 ∧ m0 / VFS_PAGE_SIZE ∈ content0

Heuristic symbolic execution applied rule assign left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkinode(m, md, boolvar, m0, m1, m2).ino = m; (s)% ⊦ files[ino, af][ino0] = (ino0 = ino ∧ ino ∈ files ⊃ af;files[ino0]); (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).size = m2; (s)% ⊦ mkfile(m, md, m0, content).size:= m1 = mkfile(m1, md, m0, content); (s)% ⊦ files[ino, af][ino, af0] = files[ino, af0]; (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).meta = md; (s)% ⊦ mkfile(m, md, m0, content).meta:= md0 = mkfile(m, md0, m0, content); (s)%⊦ mkfile(m, md, m0, content) = mkfile(m1, md0, m2, content0) ↔ m = m1 ∧ md = md0 ∧ m0 = m2 ∧ content = content0; (ws)% ⊦ ino ∈ files[ino0, af] ↔ ino ∈ files; (s)