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

σ0 = ⌜x0⌝ , ⌜x⌝ = (σ to n0 + m) + σ0 ⊦ ⌜n0⌝ + # σ0 = ⌜0⌝ , ⌜((⌜n0⌝ + ⌜m⌝ ) + # σ0).n - m⌝ = #((σ to m + n0) from m) + # σ0, # σ < ⌜m⌝ + ⌜n0⌝ , # σ < ⌜n0⌝ + ⌜m⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
⌜x⌝ = (σ to n0 + m) + ⌜x0⌝ ⊦ # x0 + n0 = 0, ⌜n0⌝ = #((σ to m + n0) from m), # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

Additional information

Used simplifier rules: size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;% ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s)% ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; (s)% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%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)% ⊦ ⌜m⌝ .n = m; (s)%n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s)%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; (s)%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf