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

abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files) ⊦ p + str ' ∈ tree ↔ m ∈ dirs ∧ str ∈ dirs[m].entries

User applied rule insert rewrite lemma

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) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files))
Proof step 5:
abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files) ⊦ ispath(p + str ', ino, dirs, files) ↔ m ∈ dirs ∧ str ∈ dirs[m].entries

Rule argument

spec: <toplevel> name: path-in-eq seq: ⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files)) subst: [ino, dirs, files, ofh, tree, p] → [ino, dirs, files, ofh, tree, p + str '] flags: <all paths>

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));% ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]; (s)% ⊦ ispath(p, n, m, dirs, files) → ispath(p, n, dirs, files); (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path] Used terms: ino, dirs, files, ofh, tree, p + str '