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Proof step 55 from proof swap-to-from-app (previous step)

⌜x⌝ = (σ to n0 + m) + ⌜x0⌝ ⊦ # x0 + n0 = 0, ⌜n0⌝ = #((σ to m + n0) from m), # σ < ⌜m⌝ + ⌜n0⌝ , # σ < ⌜n0⌝ + ⌜m⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: fin-size-from seq: ⊦ ⌜n⌝ < # σ ∧ finp(σ) → #(σ from n) = ⌜(# σ).n - n⌝ subst: [n, σ] → [m, σ to m + n0] flags: rewrite, current

Additional information

Used simplifier rules: size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%fin-to: ⊦ finp(σ to m);%size-empty-equiv: ⊦ # σ = ⌜0⌝ ↔ σ = [];%ton-fromn: ⊦ ¬ # σ < ⌜n⌝ → (σ to n) from n = [];%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%prime-add-fin: n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; from specification nat-inf%sel: ⊦ ⌜m⌝ .n = m; from specification nat-inf-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%eq-01: ⊦ m + n - m = n; from specification nat%inj: ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; from specification nat-inf-basic%ls-natinf-to-nat-1-2: ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; from specification nat-inf%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%add-base: ⊦ n + 0 = n; from specification nat-basic2%add-all-zero: ⊦ n∞ + ⌜0⌝ = n∞; from specification nat-inf Used terms: m, σ to m + n0