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Proof step 3 from proof abs-of-put (previous step)

⊦ ∀ 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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ fd ∈ abs-of(of) ↔ fd ∈ of
Proof step 5:
⊦ (fd0 ∈ abs-of(of)[fd, mkfh(⌜ino⌝ , mode, n)] ↔ fd0 ∈ 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

Additional information

Used simplifier rules: abs-of-in: ⊦ fd ∈ abs-of(of) ↔ fd ∈ of;