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Proof step 3 from proof abs-fold-del (previous step)

tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(# entries - 1, md, m0, m1, entries -- str) * abs(sdt -- str, entries -- str)))(dirs), p ∈ tree, str ∈ entries ⊦ abs(tree -- (p + str '), ROOT_INO)(dirs)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# entries = m + 1, tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(m, md, m0, m1, entries -- str) * abs(sdt -- str, entries -- str)))(dirs), str ∈ entries, p ∈ tree ⊦ # entries = 0, abs(tree -- (p + str '), ROOT_INO)(dirs)

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# entries, m]

Additional information

Used simplifier rules: ⊦ # entries = 0 ↔ entries = ∅; (s)% ⊦ # ∅ = 0; (s)% ⊦ ∅ -- str = ∅; (s)% ⊦ abs(sdt, ∅) = (⌜sdt = ∅⌝ ∧ emp); (s)% ⊦ (⌜bv⌝ ∧ D) * D0 = (⌜bv⌝ ∧ D * D0); (s)% ⊦ emp * D = D; (s)%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]%and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]% ⊦ ((D ∧ D0))(dirs) ↔ D(dirs) ∧ D0(dirs); (s)% ⊦ ⌜bv⌝ (dirs) ↔ bv; (s)% ⊦ ¬ str ∈ ∅; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # entries, m