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Proof step 1 from proof abs-of-put (conclusion)

⊦ abs-of(of)[fd, mkfh(⌜ino⌝ , mode, n)] = abs-of(of[fd, mkfile(ino, mode, n)])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ofh = ofh0 ↔ (∀ fd. (fd ∈ ofh ↔ fd ∈ ofh0) ∧ (fd ∈ ofh → ofh[fd] = ofh0[fd]))
Proof step 3:
⊦ ∀ fd0. (fd0 ∈ abs-of(of)[fd, mkfh(⌜ino⌝ , mode, n)] ↔ fd0 ∈ abs-of(of[fd, mkfile(ino, mode, n)])) ∧ (fd0 ∈ abs-of(of)[fd, mkfh(⌜ino⌝ , mode, n)] → abs-of(of)[fd, mkfh(⌜ino⌝ , mode, n)][fd0] = abs-of(of[fd, mkfile(ino, mode, n)])[fd0])

Rule argument

spec: store, inst: open-file-handles name: k-Extension seq: ⊦ ofh = ofh0 ↔ (∀ fd. (fd ∈ ofh ↔ fd ∈ ofh0) ∧ (fd ∈ ofh → ofh[fd] = ofh0[fd])) subst: [ofh0, ofh] → [abs-of(of[fd, mkfile(ino, mode, n)]), abs-of(of)[fd, mkfh(⌜ino⌝ , mode, n)]] flags: <all paths>

Additional information

Used terms: abs-of(of[fd, mkfile(ino, mode, n)]), abs-of(of)[fd, mkfh(⌜ino⌝ , mode, n)]