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

⊦ ∀ fid. (fid ∈ abs-files(files) -- ⌜ino⌝ ↔ fid ∈ abs-files(files -- ino)) ∧ (fid ∈ abs-files(files) -- ⌜ino⌝ → (abs-files(files) -- ⌜ino⌝ )[fid] = abs-files(files -- ino)[fid])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (fid ≠ ⌜ino⌝ ∧ fid ∈ abs-files(files) ↔ fid ∈ abs-files(files -- ino)) ∧ (fid ≠ ⌜ino⌝ ∧ fid ∈ abs-files(files) → abs-files(files)[fid] = abs-files(files -- ino)[fid])

Rule argument

Additional information

Used simplifier rules: ⊦ fid ≠ fid0 → (fs -- fid0)[fid] = fs[fid]; (s)% ⊦ fid ∈ fs -- fid0 ↔ fid ≠ fid0 ∧ fid ∈ fs; (s)