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Proof step 53 from proof swap-to-from-app (previous step)

⌜m⌝ < #((σ to n0 + m) + σ0), finp((σ to n0 + m) + σ0) ⊦ ⌜(#((σ to n0 + m) + σ0)).n - m⌝ = #((σ to m + n0) from m) + # σ0, # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 54:
⌜x⌝ = (σ to n0 + m) + σ0 ⊦ ⌜n0⌝ + # σ0 = ⌜0⌝ , ⌜((⌜n0⌝ + ⌜m⌝ ) + # σ0).n - m⌝ = #((σ to m + n0) from m) + # σ0, # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

Additional information

Used simplifier rules: size-append: ⊦ #(σ + σ1) = # σ + # σ1;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;% ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; (s)% ⊦ finp(⌜x⌝ ); (s)%stream-fin-forward: ⊦ finp(σ) → (∃ x. ⌜x⌝ = σ); (forward) from specification stream-basic%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ ⌜n⌝ < n∞ + ⌜n⌝ ↔ n∞ ≠ ⌜0⌝ ; (s)%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)