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Proof step 3 from proof from-infto-exchange (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (σ to m0 + m) from m = (σ from m) to m0, # σ < ⌜m⌝ + ⌜m0⌝

Rule argument

Additional information

Used simplifier rules: toinf-add: ⊦ σ to∞ ⌜n⌝ + ⌜m⌝ = σ to n + m;%toinf-fin: ⊦ σ to∞ ⌜n⌝ = σ to n;%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