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

⊦ σ from m from n = σ from m + n, # σ < ⌜m + n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ σ from m from n = σ from m + n, # σ < ⌜m⌝ + ⌜n⌝

Rule argument

Additional information

Used simplifier rules: 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)