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Proof step 12 from proof abs-files-new (previous step)

⊦ 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])

User applied rule constructor cut

And obtained the following new goals

Proof step 13:
fid = ⌜ino0⌝ ⊦ 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

var: fid

Additional information

Used simplifier rules: fid freely generated by ⌜ ⌝;;