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

ispath(q, n, m, dirs, files), p ⊑ q ⊦ ispath(p, n, dirs, files)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ p ⊑ p0 ↔ (∃ p1. p + p1 = p0)

Rule argument

spec: list-last, inst: path name: prefix seq: ⊦ p ⊑ p0 ↔ (∃ p1. p + p1 = p0) subst: [p0, p] → [q, p] flags: <all paths>

Additional information

Used simplifier rules: path-expath-prefix: ⊦ ispath(p + q, ino, m, dirs, files) → ispath(p, ino, dirs, files);%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path] Used terms: q, p