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Proof step 11 from proof files-del-eq (previous step)

ino ∈ files1, ino ∈ files0, ino0 ≠ ino → (ino0 ∈ files0 ↔ ino0 ∈ files1), ino0 ≠ ino ∧ ino0 ∈ files0 → files0[ino0] = files1[ino0], ∀ ino1. (ino1 ≠ ino → (ino1 ∈ files0 ↔ ino1 ∈ files1)) ∧ (ino1 ≠ ino ∧ ino1 ∈ files0 → files0[ino1] = files1[ino1]) ⊦ 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 case distinction

And closed the goal!

Rule argument

pos: right 2

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