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Proof step 2 from proof from-infto-exchange (previous step)

⊦ (σ to∞ ⌜m⌝ + n∞) from m = (σ from m) to∞ n∞, # σ < ⌜m⌝ + n∞

User applied rule constructor cut

And obtained the following new goals

Proof step 3:
n∞ = ⌜m0⌝ ⊦ (σ to∞ ⌜m⌝ + n∞) from m = (σ from m) to∞ n∞, # σ < ⌜m⌝ + n∞

Rule argument

var: n∞

Additional information

Used simplifier rules: toinf-inf: ⊦ σ to∞ ∞ = σ;%natinf freely generated by ∞, ⌜ ⌝;;%add-inf2: ⊦ ⌜m⌝ + ∞ = ∞; from specification nat-inf%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