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Proof step 35 from proof dups-eqsize-entries (previous step)

dentrykey(m, str0) = dentrykey(m, str) ∨ dentrykey(m, str0) ∈ ks0 ↔ str0 ∈ entries0[str, ino], key ∈ ks0 → key.dentry? ∧ key.ino = m, ∀ key. key.dentry? ∧ key.ino = m → (key = dentrykey(key.ino, str) ∨ key ∈ ks0 ↔ key.name ∈ entries0[str, ino]), ∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = m, ∀ ks1, ino, entries. # ks1 < #(ks0 ++ dentrykey(m, str)) → (∀ key. key ∈ ks1 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks1 ↔ key.name ∈ entries)) → # ks1 = # entries ⊦ # ks0 = # entries0, dentrykey(m, str0) ∈ ks0 ↔ str0 ∈ entries0, str ∈ entries0, dentrykey(m, str) ∈ ks0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: ⊦ str ∈ entries[str0, ino] ↔ str = str0 ∨ str ∈ entries; (ws)% ⊦ dentrykey(m, str) = dentrykey(m0, str0) ↔ m = m0 ∧ str = str0; (ws)