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Proof step 24 from proof dups-eqsize-entries (previous step)

str ∈ entries, ∀ key0. key0 ∈ ks0 → key0.dentry? ∧ key0.ino = m, ∀ key. key.dentry? ∧ key.ino = m → (key = dentrykey(key.ino, str) ∨ key ∈ ks0 ↔ key.name ∈ entries), ∀ 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 + 1 = # entries, dentrykey(m, str) ∈ ks0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ str ∈ entries ↔ (∃ ino, entries0. entries = entries0[str, ino] ∧ ¬ str ∈ entries0)
Proof step 26:
∀ key. key.dentry? ∧ key.ino = m → (key = dentrykey(key.ino, str) ∨ key ∈ ks0 ↔ key.name ∈ entries), ∀ key0. key0 ∈ ks0 → key0.dentry? ∧ key0.ino = m, ∃ ino, entries0. entries = entries0[str, ino] ∧ ¬ str ∈ entries0, ∀ 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 + 1 = # entries, dentrykey(m, str) ∈ ks0

Rule argument

spec: store, inst: dir-entries name: exin seq: ⊦ str ∈ entries ↔ (∃ ino, entries0. entries = entries0[str, ino] ∧ ¬ str ∈ entries0) subst: [entries, str] → [entries, str] flags: <all paths>

Additional information

Used terms: entries, str