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Proof step 22 from proof from-append-add (previous step)

⊦ # σ < ⌜n⌝ , #((σ to n) + ⌜nf⌝ from n1 + n) = ∞ ∧ ((σ to n) + ⌜nf⌝ from n1 + n)[n0] = nf(n0 + n1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ ⌜n⌝ < # σ ∧ finp(σ) → #(σ from n) = ⌜(# σ).n - n⌝
Proof step 24:
⊦ # σ < ⌜n⌝ , #((σ to n) + ⌜nf⌝ from n1 + n) = ∞ ∧ ((σ to n) + ⌜nf⌝ from n1 + n)[n0] = nf(n0 + n1), ⌜n1 + n⌝ < #((σ to n) + ⌜nf⌝ ) ∧ finp((σ to n) + ⌜nf⌝ )

Rule argument

spec: <toplevel> name: fin-size-from seq: ⊦ ⌜n⌝ < # σ ∧ finp(σ) → #(σ from n) = ⌜(# σ).n - n⌝ subst: [n, σ] → [n1 + n, (σ to n) + ⌜nf⌝ ] flags: <all paths>

Additional information

Used simplifier rules: size-append: ⊦ #(σ + σ1) = # σ + # σ1;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%size-inf: ⊦ # ⌜nf⌝ = ∞;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%neq-fin-inf-01: ⊦ ⌜x⌝ ≠ σ + ⌜nf0⌝ ;% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ∞ + ⌜m⌝ = ∞; (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)% ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; (s)% ⊦ ⌜m⌝ ≠ ∞; (s)%stream-fin-forward: ⊦ finp(σ) → (∃ x. ⌜x⌝ = σ); (forward) from specification stream-basic Used terms: n1 + n, (σ to n) + ⌜nf⌝