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Proof step 3 from proof files-del-eq (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 4:
files0 -- ino = files1 -- ino, ino ∈ files0, ino ∈ files1 ⊦ abs-files(files0)[⌜ino⌝ , f] = abs-files(files), files1 = files0[ino, files1[ino]]

Rule argument

pos: right 2 terms: files1[ino] abort: F discard: T computed: F

Additional information

Used terms: files1[ino]