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

∀ ino. (ino ∈ files0 ↔ ino ∈ files1) ∧ ( ino ∈ files0 → files0[ino].size = files1[ino].size ∧ files0[ino].meta = files1[ino].meta ∧ files0[ino].nlink = files1[ino].nlink ∧ files0[ino].content.stream to files0[ino].size = files1[ino].content.stream to files1[ino].size) ⊦ abs-files(files0) = abs-files(files1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ fs = fs0 ↔ (∀ fid. (fid ∈ fs ↔ fid ∈ fs0) ∧ (fid ∈ fs → fs[fid] = fs0[fid]))
Proof step 7:
∀ ino. (ino ∈ files0 ↔ ino ∈ files1) ∧ ( ino ∈ files0 → files0[ino].size = files1[ino].size ∧ files0[ino].meta = files1[ino].meta ∧ files0[ino].nlink = files1[ino].nlink ∧ files0[ino].content.stream to files0[ino].size = files1[ino].content.stream to files1[ino].size) ⊦ ∀ fid. (fid ∈ abs-files(files0) ↔ fid ∈ abs-files(files1)) ∧ (fid ∈ abs-files(files0) → abs-files(files0)[fid] = abs-files(files1)[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(files1), abs-files(files0)] flags: <all paths>

Additional information

Used terms: abs-files(files1), abs-files(files0)