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Proof step 7 from proof files-del-eq (previous step)

files0 -- ino = files1 -- ino, ino ∈ files0, ino ∈ files1 ⊦ abs-files(files0)[⌜ino⌝ , f] = abs-files(files), (ino0 ∈ files1 ↔ ino0 ∈ files0) ∧ (ino0 ∈ files1 → files1[ino0] = (ino0 = ino ⊃ files1[ino];files0[ino0]))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ files = files0 ↔ (∀ ino. (ino ∈ files ↔ ino ∈ files0) ∧ (ino ∈ files → files[ino] = files0[ino]))
Proof step 9:
ino ∈ files1, ino ∈ files0, ∀ ino0. (ino0 ∈ files0 -- ino ↔ ino0 ∈ files1 -- ino) ∧ (ino0 ∈ files0 -- ino → (files0 -- ino)[ino0] = (files1 -- ino)[ino0]) ⊦ abs-files(files0)[⌜ino⌝ , f] = abs-files(files), (ino0 ∈ files1 ↔ ino0 ∈ files0) ∧ (ino0 ∈ files1 → files1[ino0] = (ino0 = ino ⊃ files1[ino];files0[ino0]))

Rule argument

spec: heap, inst: file-heap name: ext seq: ⊦ files = files0 ↔ (∀ ino. (ino ∈ files ↔ ino ∈ files0) ∧ (ino ∈ files → files[ino] = files0[ino])) subst: [files0, files] → [files1 -- ino, files0 -- ino] flags: <all paths>

Additional information

Used terms: files1 -- ino, files0 -- ino