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Proof step 3 from proof abs-has-rootpath (previous step)

abs(tree, ino0)(dirs), ino1 ∈ dirs ⊦ ∃ p. ispath(p, ino0, ino1, dirs, files)

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ ((abs(tree, ino0) * {dirs0} ))(dirs) ∧ ino1 ∈ dirs ∧ ¬ ino1 ∈ dirs0 → (∃ p. ispath(p, ino0, ino1, dirs, files))

Rule argument

spec: <toplevel> name: abs-has-rootpath-h seq: ⊦ ((abs(tree, ino0) * {dirs0} ))(dirs) ∧ ino1 ∈ dirs ∧ ¬ ino1 ∈ dirs0 → (∃ p. ispath(p, ino0, ino1, dirs, files)) subst: [tree, ino0, dirs0, dirs, ino1, files] → [tree, ino0, ∅, dirs, ino1, files]

Additional information

Used simplifier rules: ⊦ {∅} = emp; (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]% ⊦ ¬ ino ∈ ∅; (s) Used terms: tree, ino0, ∅, dirs, ino1, files