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Proof step 9 from proof files-del-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
ino ∈ files0, ino ∈ files1, ∀ 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]))

Rule argument

Additional information

Used simplifier rules: ino ∈ files ⊦ ino ≠ ino0 → (files -- ino0)[ino] = files[ino]; (s)% ⊦ ino ∈ files -- ino0 ↔ ino ≠ ino0 ∧ ino ∈ files; (s)