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Proof step 45 from proof swap-to-from-app (previous step)

#((σ to m + n0) from m) = ⌜n1⌝ , #((σ to n0 + m) + σ0 from m) = ⌜n1⌝ + # σ0, n2 + n1 = n3 + n0, ⌜n2⌝ + ⌜n1⌝ < ⌜n0⌝ + # σ0, ⌜n2⌝ < # σ0, ⌜n1⌝ < ⌜n0⌝ + # σ0, n0 ≤ n1 + n2, n0 ≤ n2 + n1 ⊦ σ0[n3] = σ0[n2], σ0 = [], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
#((σ to m + n0) from m) = ⌜n1⌝ , #((σ to n0 + m) + σ0 from m) = ⌜n1⌝ + # σ0, n2 + n1 = n3 + n0, ⌜n2⌝ + ⌜n1⌝ < ⌜n0⌝ + # σ0, ⌜n2⌝ < # σ0, ⌜n1⌝ < ⌜n0⌝ + # σ0 ⊦ σ0[n3] = σ0[n2], σ0 = [], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)%c: ⊦ m + n = n + m; from specification nat-basic