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Proof step 1 from proof of (conclusion)

⊦ ⌜ino⌝ ∈ abs-of(of).fids ↔ is-open(ino, of)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ fid ∈ ofh.fids ↔ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid)
Proof step 3:
⊦ (∃ n. n ∈ abs-of(of) ∧ abs-of(of)[n].fid = ⌜ino⌝ ) ↔ is-open(ino, of)

Rule argument

spec: posix-base name: ofh-fids seq: ⊦ fid ∈ ofh.fids ↔ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid) subst: [ofh, fid] → [abs-of(of), ⌜ino⌝ ] flags: <all paths>

Additional information

Used terms: abs-of(of), ⌜ino⌝