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Proof step 3 from proof abs-files-new (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ fs = fs0 ↔ (∀ fid. (fid ∈ fs ↔ fid ∈ fs0) ∧ (fid ∈ fs → fs[fid] = fs0[fid]))
Proof step 5:
⊦ 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])

Rule argument

spec: store, inst: file-data-store name: k-Extension seq: ⊦ fs = fs0 ↔ (∀ fid. (fid ∈ fs ↔ fid ∈ fs0) ∧ (fid ∈ fs → fs[fid] = fs0[fid])) subst: [fs0, fs] → [abs-files((files ++ ino)[ino, mkfile(0, md, 1, ∅)]), abs-files(files)[⌜ino⌝ , mkfile(md, mkbuf(0))]] flags: <all paths>

Additional information

Used terms: abs-files((files ++ ino)[ino, mkfile(0, md, 1, ∅)]), abs-files(files)[⌜ino⌝ , mkfile(md, mkbuf(0))]