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Proof step 27 from proof path-unfold (previous step)

(((abs|(tree, ino0, p, ino) * abs(tree[p], ino)) * D))(dirs), p ∈ tree ⊦ ((abs(tree, ino0) * D))(dirs)

User applied rule structural induction

And obtained the following new goals

Proof step 28:
⊦ abs|(tree, ino, ε, ino0) = (emp ∧ ⌜ino = ino0⌝ )
Proof step 29:
(((abs|(tree, ino0, str + p0, ino) * abs(tree[str + p0], ino)) * D))(dirs), str + p0 ∈ tree, ∀ tree, ino0, ino, D, dirs. (((abs|(tree, ino0, p0, ino) * abs(tree[p0], ino)) * D))(dirs) ∧ p0 ∈ tree → ((abs(tree, ino0) * D))(dirs) ⊦ ((abs(tree, ino0) * D))(dirs)

Rule argument

indvar: p

Additional information

Used simplifier rules: absc-emp: ⊦ abs|(tree, ino, ε, ino0) = (emp ∧ ⌜ino = ino0⌝ );%path freely generated by ε, +;;% ⊦ ε ∈ tree; (s)% ⊦ tree[ε] = tree; (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)% ⊦ ⌜false⌝ = false; (s)% ⊦ ¬ false(dirs); (s)