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Proof step 144 from proof afs-cons-rename-dir-negdentry (previous step)

(entries2[str] = m0 ⊃ mkdir(m4 + 1, md1, # links(entries2[str], dirs0), m7 + 1, entries0[str0, entries2[str]]) ;(entries2[str] = m ⊃ mkdir(m1, md0, # links(entries2[str], dirs0), m2, entries2 -- str) ;mkdir(# entries1, md4, # links(entries2[str], dirs0), # subdirs(entries2[str], dirs0), entries1) ) ) = mkdir(m15, md7, m16, m17, entries4), dirs0[m0] = mkdir(m4, md5, # links(m0, dirs0), m7, entries0), dirs0[m] = mkdir(m1 + 1, md, # links(m, dirs0), m2 + 1, entries2), files[entries2[str]] = mkfile(m9, md2, m11, content), dirs0[entries2[str]] = mkdir(# entries1, md4, # links(entries2[str], dirs0), # subdirs(entries2[str], dirs0), entries1), afs-cons(dirs0, files), ROOT_INO ∈ dirs0, entries2[str] ∈ dirs0, m ∈ dirs0, m0 ∈ dirs0, str ∈ entries2, # links(entries2[str], dirs0) ≤ 1, name ∈ entries1 → entries1[name] ∈ dirs0 ∨ entries1[name] ∈ files, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ # links(entries2[str], dirs0) - 1 = 0, links(entries2[str], dirs0) = ∅ ++ mklink(m, str), m = m0, str0 ∈ entries0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# links(entries2[str], dirs0), m3]

Additional information

Used simplifier rules: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); (s)%afs-cons-dir-nlink: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) → # links(ino, dirs) = n; (forward) from specification afs-base% ⊦ n ≠ 0 → (m + n = 1 ↔ n = 1 ∧ m = 0); (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%size-zero-02: ⊦ # ls = 0 → ls = ∅; (forward) from specification set-basic[alink-set]%⊦ p_ino ∈ dirs ∧ dirs[p_ino] = mkdir(m, md, n, n0, entries) ∧ str ∈ entries → links(entries[str], dirs) ≠ ∅; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # links(entries2[str], dirs0), m3