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Proof step 3 from proof swap-to-from (previous step)

n = n0 + m, m ≤ n ⊦ (σ to n) from m = (σ from m) to n0, # σ < ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (σ to n0 + m) from m = (σ from m) to n0, # σ < ⌜n0⌝ + ⌜m⌝

Rule argument

Additional information

Used simplifier rules: add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat