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Proof step 20 from proof from-append-add (previous step)

σ0 = ⌜nf⌝ ⊦ # σ0 < ⌜n1⌝ , # σ < ⌜n⌝ , #((σ to n) + σ0 from n1 + n) = #(σ0 from n1) ∧ (⌜n0⌝ < #((σ to n) + σ0 from n1 + n) → ((σ to n) + σ0 from n1 + n)[n0] = (σ0 from n1)[n0]), ⌜n1⌝ < # σ0 ∧ finp(σ0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: size-inf: ⊦ # ⌜nf⌝ = ∞;%from-inf: ⊦ ⌜nf⌝ from n = ⌜nf from n⌝ ;%size-append: ⊦ #(σ + σ1) = # σ + # σ1;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%less-size-from: ⌜n⌝ < # σ ⊦ ⌜n0⌝ < #(σ from n) ↔ ⌜n0 + n⌝ < # σ;%sel-inf: ⊦ ⌜nf⌝ [n] = nf(n);%get-from-inf: ⊦ ((nf from n))(m) = nf(m + n);%test-01: ⊦ ¬ finp(⌜nf⌝ ); from specification stream-basic%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%addeqinf-cancel: ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; from specification nat-inf%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-inf2: ⊦ ⌜m⌝ + ∞ = ∞; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%less-inf1: ⊦ ¬ ∞ < n∞; from specification nat-inf