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Proof step 2 from proof path-in (previous step)

abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files) ⊦ p ∈ tree ∧ (tree[p].dir? ↔ m ∈ dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files))
Proof step 4:
abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files) ⊦ tree[p].dir? ↔ m ∈ dirs

Rule argument

Additional information

Used simplifier rules: path-in-eq: ⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files));% ⊦ ispath(p, n, m, dirs, files) → ispath(p, n, dirs, files); (s)