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Proof step 15 from proof replay-orphans-trunc (previous step)

datakey(m, m0) ∈ ri, key ∈ ro, datakey(m, m0) ∈ ri \ datakeys(key, n, ri), ¬ key.inode? ∨ m ≠ key.ino ∨ ¬ n ≤ m0 * VFS_PAGE_SIZE ⊦ # ro = 0, (ri \ datakeys(key, n, ri))[datakey(m, m0)] = ri[datakey(m, m0)], ro = ∅, inodekey(m) ∈ ro, datakey(m, m0) ∈ ro

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
key ∈ ro, datakey(m, m0) ∈ ri, ¬ key.inode? ∨ m ≠ key.ino ∨ ¬ n ≤ m0 * VFS_PAGE_SIZE ⊦ # ro = 0, (ri \ datakeys(key, n, ri))[datakey(m, m0)] = ri[datakey(m, m0)], ro = ∅, datakey(m, m0) ∈ ro, inodekey(m) ∈ ro

Rule argument

Additional information

Used simplifier rules: ⊦ datakey(m, m0).part = m0; (s)% ⊦ datakey(m, m0).ino = m; (s)% ⊦ datakey(m, m0).data?; (s)%⊦ 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)