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Proof step 11 from proof abs-files-new (previous step)

⊦ fid = ⌜ino⌝ , ino = 0, ino ∈ files, (fid ∈ abs-files(files)[⌜ino⌝ , mkfile(md, mkbuf(0))] ↔ fid ∈ abs-files((files ++ ino)[ino, mkfile(0, md, 1, ∅)])) ∧ ( fid ∈ abs-files(files)[⌜ino⌝ , mkfile(md, mkbuf(0))] → abs-files(files)[⌜ino⌝ , mkfile(md, mkbuf(0))][fid] = abs-files((files ++ ino)[ino, mkfile(0, md, 1, ∅)])[fid])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ fid = ⌜ino⌝ , ino = 0, ino ∈ files, (fid ∈ abs-files(files) ↔ fid ∈ abs-files((files ++ ino)[ino, mkfile(0, md, 1, ∅)])) ∧ (fid ∈ abs-files(files) → abs-files(files)[fid] = abs-files((files ++ ino)[ino, mkfile(0, md, 1, ∅)])[fid])

Rule argument

Additional information

Used simplifier rules: ⊦ fid ≠ fid0 → fs[fid0, f][fid] = fs[fid]; (s)% ⊦ fid ≠ fid0 → (fid ∈ fs[fid0, f] ↔ fid ∈ fs); (s)