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Proof step 5 from proof from-to-exchange (previous step)

(tl(σ) to m0 + n) from m0 = (tl(σ) from m0) to n ∨ # tl(σ) < ⌜m0⌝ + ⌜n⌝ , ∀ σ, n. (σ to m0 + n) from m0 = (σ from m0) to n ∨ # σ < ⌜m0⌝ + ⌜n⌝ ⊦ (σ to (m0 + 1) + n) from m0 + 1 = (σ from m0 + 1) to n, # σ < (⌜m0⌝ + ⌜1⌝ ) + ⌜n⌝

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: stream inst: name: elim-hd-tl seq: ⊦ σ ≠ [] → (a = hd(σ) ∧ σ0 = tl(σ) ↔ σ = a + σ0) subst: [σ, a, σ0] → [σ, a, σ0] flags: rewrite, current

Additional information

Used simplifier rules: size-empty: ⊦ # [] = ⌜0⌝ ;%cons-append: ⊦ a + σ = ⌜a '⌝ + σ;%to-succ-cons: ⊦ ⌜a '⌝ + σ to n + 1 = a + (σ to n);%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%from-succ-cons: ⊦ ⌜a '⌝ + σ from n + 1 = σ from n;%size-append: ⊦ #(σ + σ1) = # σ + # σ1;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%len-02: ⊦ # a ' = 0 +1; from specification list%add-ls-cancel-num-04: ⊦ m ≤ n → (⌜(* m)⌝ + n∞2 < ⌜(* n)⌝ + n∞1 ↔ n∞2 < ⌜n - m⌝ + n∞1); from specification nat-inf%add-all-zero: ⊦ n∞ + ⌜0⌝ = n∞; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%zerols-neqzero: ⊦ ⌜0⌝ < n∞ ↔ n∞ ≠ ⌜0⌝ ; from specification nat-inf%add-cancel-num-02: ⊦ n < m → (⌜(* n)⌝ = ⌜(* m)⌝ + n∞2 ↔ false); from specification nat-inf Used terms: σ, a, σ0