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Proof step 5 from proof abs-files-new (previous step)

⊦ ino = 0, ino ∈ files, ∀ fid. (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 6:
⊦ 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])

Proof step 5 from proof abs-files-new (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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