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Proof step 19 from proof dups-eqsize-ls (previous step)

∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls), ∀ key. key ∈ ks → key.dentry?, ∃ ks0. ks = ks0 ++ key ∧ ¬ key ∈ ks0, ∀ ls0, ks. # ls0 < # ls → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls0)) → # ks = # ls0 ⊦ # ks = # ls, # ks = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
∀ key0. key0 = key ∨ key0 ∈ ks0 → key0.dentry?, ∀ key0. key0.dentry? → (key0 = key ∨ key0 ∈ ks0 ↔ mklink(key0.ino, key0.name) ∈ ls), ∀ ls0, ks. # ls0 < # ls → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls0)) → # ks = # ls0 ⊦ # ks0 + 1 = # ls, key ∈ ks0

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ ks ++ key0 ↔ key = key0 ∨ key ∈ ks; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ ¬ key ∈ ks → #(ks ++ key) = # ks + 1; (s)