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Proof step 26 from proof dups-eqsize-entries (previous step)

∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
∀ key0. key0 ∈ ks0 → key0.dentry? ∧ key0.ino = m, ∀ key. key.dentry? ∧ key.ino = m → (key = dentrykey(key.ino, str) ∨ key ∈ ks0 ↔ key.name ∈ entries0[str, ino]), ∀ 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, str) ∈ ks0, str ∈ entries0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ ¬ str ∈ entries → # entries[str, ino] = # entries + 1; (s)