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Proof step 27 from proof dups-eqsize-entries (previous step)

∀ 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

User applied rule apply induction

And obtained the following new goals

Proof step 28:
∀ key. key.dentry? ∧ key.ino = m → (key = dentrykey(key.ino, str) ∨ key ∈ ks0 ↔ key.name ∈ entries0[str, ino]), ∀ 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 = # entries0, str ∈ entries0, dentrykey(m, str) ∈ ks0, (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = m) ∧ (∀ key. key.dentry? ∧ key.ino = m → (key ∈ ks0 ↔ key.name ∈ entries0))

Rule argument

subst: [ks1, ino, entries] → [ks0, m, entries0]

Additional information

Used simplifier rules: ⊦ ¬ key ∈ ks → #(ks ++ key) = # ks + 1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: ks0, m, entries0