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Proof step 32 from proof dups-eqsize-ls (previous step)

key ∈ ks0, ∀ key. key.dentry? → (key = dentrykey(m, str) ∨ key ∈ ks0 ↔ mklink(key.ino, key.name) = mklink(m, str) ∨ mklink(key.ino, key.name) ∈ ls0), ∀ key0. key0 ∈ ks0 → key0.dentry?, ∀ ls1, ks. # ls1 < #(ls0 ++ mklink(m, str)) → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls1)) → # ks = # ls1 ⊦ # ks0 = # ls0, key.dentry?, mklink(m, str) ∈ ls0, dentrykey(m, str) ∈ ks0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
key ∈ ks0, ∀ key. key.dentry? → (key = dentrykey(m, str) ∨ key ∈ ks0 ↔ mklink(key.ino, key.name) = mklink(m, str) ∨ mklink(key.ino, key.name) ∈ ls0), ∀ key0. key0 ∈ ks0 → key0.dentry?, ∀ ls1, ks. # ls1 < #(ls0 ++ mklink(m, str)) → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls1)) → # ks = # ls1 ⊦ # ks0 = # ls0, # ks0 = 0, ks0 = ∅, key.dentry?, mklink(m, str) ∈ ls0, dentrykey(m, str) ∈ ks0

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ ks ≠ ∅ → # ks ≠ 0; (forward) from specification set-basic[node-index]%ne-01: ⊦ key ∈ ks → ks ≠ ∅; (forward) from specification set-basic[node-index]