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Proof step 6 from proof files-del-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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]))

Rule argument

Additional information

Used simplifier rules: ⊦ files[ino, af][ino0] = (ino0 = ino ∧ ino ∈ files ⊃ af;files[ino0]); (s)% ⊦ ino ∈ files[ino0, af] ↔ ino ∈ files; (s)