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Proof step 2 from proof ispath-target (previous step)

ispath(p, ino, dirs, files) ⊦ ∃ m. ispath(p, ino, m, dirs, files)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ispath(p, n, dirs, files) ↔ (∃ m. ispath(p, n, m, dirs, files))
Proof step 4:
∃ m. ispath(p, ino, m, dirs, files) ⊦ ∃ m. ispath(p, ino, m, dirs, files)

Rule argument

spec: afs-base name: expath seq: ⊦ ispath(p, n, dirs, files) ↔ (∃ m. ispath(p, n, m, dirs, files)) subst: [files, dirs, n, p] → [files, dirs, ino, p] flags: <all paths>

Additional information

Used terms: files, dirs, ino, p