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Proof step 2 from proof files-del-eq (previous step)

ino ∈ files0, ino ∈ files1 ⊦ files0 -- ino = files1 -- ino ↔ (∃ af. files1 = files0[ino, af]), abs-files(files0)[⌜ino⌝ , f] = abs-files(files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
files0 -- ino = files1 -- ino, ino ∈ files1, ino ∈ files0 ⊦ abs-files(files0)[⌜ino⌝ , f] = abs-files(files), ∃ af. files1 = files0[ino, af]

Rule argument

Additional information

Used simplifier rules: ino ∈ files ⊦ files[ino, af] -- ino = files -- ino; (s)