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Proof step 3 from proof absc-path (previous step)

((abs|(tree, ino, p, ino0) * D))(dirs), ispath(p, ino, ino1, dirs, files), p ∈ tree ⊦ ino0 = ino1

User applied rule structural induction

And obtained the following new goals

Proof step 4:
⊦ abs|(tree, ino, ε, ino0) = (emp ∧ ⌜ino = ino0⌝ )
Proof step 5:
((abs|(tree, ino, str + p0, ino0) * D))(dirs), ispath(str + p0, ino, ino1, dirs, files), str + p0 ∈ tree, ∀ tree, ino, ino0, D, dirs, ino1, files. ((abs|(tree, ino, p0, ino0) * D))(dirs) ∧ ispath(p0, ino, ino1, dirs, files) ∧ p0 ∈ tree → ino0 = ino1 ⊦ ino0 = ino1

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)% ⊦ ispath(ε, n, m, dirs, files) ↔ valid-ino(n, dirs, files) ∧ m = n; (s)% ⊦ valid-ino(ino, dirs, files) ↔ ino ∈ dirs ∨ ino ∈ files; (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)