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Proof step 133 from proof refine-writemeta (previous step)

(ino1 ∈ dirs ⊃ mkinode(ino1, dirs[ino1].meta, true, dirs[ino1].nlink, dirs[ino1].nsubdirs, dirs[ino1].size) ;mkinode(ino1, files[ino1].meta, false, files[ino1].nlink, 0, files[ino1].size) ) = mkinode(m, md0, boolvar, m0, m1, m2), root[path].dir? ↔ ino1 ∈ dirs, files[m] = mkfile(m3, md, m4, content), dirs[m] = mkdir(m5, md1, m6, m7, entries), root.dir?, dom of < ino0, abs(root, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(root, abs-files(files), abs-of(of)), ispath(path, ROOT_INO, ino1, dirs, files), of-cons(of, dirs, files), pw(user, md0), px-parent(user, path, ROOT_INO, dirs, files), psynced → asynced0, ino1 ∈ dirs ∨ ino1 ∈ files ⊦ m ∈ dirs ∧ m0 = m6 ∧ m2 = m5 ∧ m1 = m7 ∧ boolvar, m ∈ files ∧ m0 = m4 ∧ m1 = 0 ∧ ¬ boolvar

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ mkinode(m, md, boolvar, m0, m1, m2) = mkinode(m3, md0, boolvar0, m4, m5, m6) ↔ m = m3 ∧ md = md0 ∧ (boolvar ↔ boolvar0) ∧ m0 = m4 ∧ m1 = m5 ∧ m2 = m6; (ws)% ⊦ mkfile(m, md, m0, content).meta = md; (s)% ⊦ mkfile(m, md, m0, content).nlink = m0; (s)% ⊦ mkdir(m, md, m0, m1, entries).meta = md; (s)% ⊦ mkdir(m, md, m0, m1, entries).nsubdirs = m1; (s)% ⊦ mkdir(m, md, m0, m1, entries).size = m; (s)% ⊦ mkdir(m, md, m0, m1, entries).nlink = m0; (s)