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

⊦ (σ to n) + ((σ from n) to n1) = σ to n + n1, # σ < ⌜n + n1⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ (σ to n) + ((σ from n) to n1) = σ to n + n1, # σ < ⌜n⌝ + ⌜n1⌝

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)