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Proof step 7 from proof replay-nonorphans-trunc (previous step)

key0 ∈ (ri ⊖ ro) \ datakeys(key, n, ri) ↔ key0 ∈ (ri \ datakeys(key, n, ri)) ⊖ ro ⊦ key ∈ ro, (key0 ∈ (ri ⊖ ro) \ datakeys(key, n, ri) ↔ key0 ∈ (ri \ datakeys(key, n, ri)) ⊖ ro) ∧ (key0 ∈ (ri ⊖ ro) \ datakeys(key, n, ri) → ((ri ⊖ ro) \ datakeys(key, n, ri))[key0] = ((ri \ datakeys(key, n, ri)) ⊖ ro)[key0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
key0 ∈ ri, key0.data? → ¬ inodekey(key0.ino) ∈ ro, ¬ key0.data? ∨ ¬ key.inode? ∨ key0.ino ≠ key.ino ∨ ¬ n ≤ key0.part * VFS_PAGE_SIZE ⊦ ((ri ⊖ ro) \ datakeys(key, n, ri))[key0] = ((ri \ datakeys(key, n, ri)) ⊖ ro)[key0], key ∈ ro, key0 ∈ ro

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ ri ⊖ ks ↔ key ∈ ri ∧ ¬ key ∈ ks ∧ (key.data? → ¬ inodekey(key.ino) ∈ ks); (ws)%⊦ key ∈ datakeys(key0, ri) ↔ key ∈ ri ∧ key.data? ∧ key0.inode? ∧ key.ino = key0.ino; (s)%⊦ key ∈ datakeys(key0, n, ri) ↔ key ∈ datakeys(key0, ri) ∧ n ≤ key.part * VFS_PAGE_SIZE; (s)% ⊦ key ∈ ri \ ks ↔ key ∈ ri ∧ ¬ key ∈ ks; (ws)