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Proof step 23 from proof dups-eqsize-entries (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
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

Rule argument

Additional information

Used simplifier rules: ⊦ dentrykey(m, str).name = str; (s)% ⊦ dentrykey(m, str).ino = m; (s)% ⊦ dentrykey(m, str).dentry?; (s)