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Proof step 24 from proof new-inode-exists (previous step)

∀ ino0. inodekey(ino0) ∈ ks0 → ino0 < ino, ∀ ks1. # ks1 < #(ks0 ++ inodekey(m)) → true → (∃ ino. ∀ ino0. inodekey(ino0) ∈ ks1 → ino0 < ino) ⊦ inodekey(m) ∈ ks0, ∀ ino0. ino0 = m ∨ inodekey(ino0) ∈ ks0 → ino0 < ino + m + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
ino + m < ino0, inodekey(ino0) ∈ ks0, ∀ ino0. inodekey(ino0) ∈ ks0 → ino0 < ino, ∀ ks1. # ks1 < #(ks0 ++ inodekey(m)) → true → (∃ ino. ∀ ino0. inodekey(ino0) ∈ ks1 → ino0 < ino) ⊦ # ks0 = 0, ks0 = ∅, inodekey(m) ∈ 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]% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)