Proof step 1 from proof toinftoinf-append-add (conclusion) (back to tree)

⊦ ¬ # σ < n∞ ∧ ¬ # σ0 < n∞1 → (σ to∞ n∞) + σ0 to∞ n∞1 + n∞ = (σ to∞ n∞) + (σ0 to∞ n∞1)

System applied rule normalize

And obtained the following new goals

Proof step 2:
(σ to∞ n∞) + σ0 to∞ n∞1 + n∞ ≠ (σ to∞ n∞) + (σ0 to∞ n∞1), ¬ # σ0 < n∞1, ¬ # σ < n∞ ⊦

Rule argument

Additional information

Proof step 2 from proof toinftoinf-append-add (previous step) (back to tree)

(σ to∞ n∞) + σ0 to∞ n∞1 + n∞ ≠ (σ to∞ n∞) + (σ0 to∞ n∞1), ¬ # σ0 < n∞1, ¬ # σ < n∞ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
(σ to∞ n∞) + σ0 to∞ n∞1 + n∞ ≠ (σ to∞ n∞) + (σ0 to∞ n∞1), ¬ # σ0 < n∞1, ¬ # σ < n∞ ⊦

Rule argument

Additional information

Proof step 3 from proof toinftoinf-append-add (previous step) (back to tree)

(σ to∞ n∞) + σ0 to∞ n∞1 + n∞ ≠ (σ to∞ n∞) + (σ0 to∞ n∞1), ¬ # σ0 < n∞1, ¬ # σ < n∞ ⊦

User applied rule constructor cut

And obtained the following new goals

Proof step 4:
n∞ = ⌜n⌝ , (σ to∞ n∞) + σ0 to∞ n∞1 + n∞ ≠ (σ to∞ n∞) + (σ0 to∞ n∞1), ¬ # σ0 < n∞1, ¬ # σ < n∞ ⊦

Rule argument

var: n∞

Additional information

Used simplifier rules: append-toinf-ls: ⊦ ¬ # σ < n∞ → σ + σ0 to∞ n∞ = σ to∞ n∞; toinf-inf: ⊦ σ to∞ ∞ = σ; append-cancel-l-02: ⊦ σ0 + σ = σ0 ↔ σ = [] ∨ ¬ finp(σ0); toinf-empty2: ⊦ ¬ # σ < n∞ → (σ to∞ n∞ = [] ↔ n∞ = ⌜0⌝ ); append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ; natinf freely generated by ∞, ⌜ ⌝;; ⊦ n∞ + ∞ = ∞; (s) 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 ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) stream-fin-forward: ⊦ finp(σ) → (∃ x. ⌜x⌝ = σ); (forward) from specification stream-basic ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 4 from proof toinftoinf-append-add (previous step) (back to tree)

n∞ = ⌜n⌝ , (σ to∞ n∞) + σ0 to∞ n∞1 + n∞ ≠ (σ to∞ n∞) + (σ0 to∞ n∞1), ¬ # σ0 < n∞1, ¬ # σ < n∞ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(σ to n) + σ0 to∞ n∞1 + ⌜n⌝ ≠ (σ to n) + (σ0 to∞ n∞1), ¬ # σ0 < n∞1, ¬ # σ < ⌜n⌝ ⊦

Rule argument

Additional information

Used simplifier rules: toinf-fin: ⊦ σ to∞ ⌜n⌝ = σ to n;

Proof step 5 from proof toinftoinf-append-add (previous step) (back to tree)

(σ to n) + σ0 to∞ n∞1 + ⌜n⌝ ≠ (σ to n) + (σ0 to∞ n∞1), ¬ # σ0 < n∞1, ¬ # σ < ⌜n⌝ ⊦

User applied rule constructor cut

And closed the goal!

Rule argument

var: n∞1

Additional information

Used simplifier rules: toinf-inf: ⊦ σ to∞ ∞ = σ; toinf-add: ⊦ σ to∞ ⌜n⌝ + ⌜m⌝ = σ to n + m; to-append-add: ⊦ ¬ # σ < ⌜n⌝ ∧ ¬ # σ0 < ⌜n1⌝ → (σ to n) + σ0 to n1 + n = (σ to n) + (σ0 to n1); append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; toinf-fin: ⊦ σ to∞ ⌜n⌝ = σ to n; natinf freely generated by ∞, ⌜ ⌝;; a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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 ⊦ ∞ + ⌜m⌝ = ∞; (s)

Proof Tree for toinftoinf-append-add (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.