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Proof step 23 from proof swap-to-from-app (previous step)

¬ ¬ ⌜n⌝ < #((σ to m + n0) from m), #((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0, ⌜n⌝ < ⌜n0⌝ + # σ0 ⊦ ((σ to n0 + m) + σ0)[m + n] = (((σ to m + n0) from m) + σ0)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
#((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m), ⌜n⌝ < #((σ to m + n0) from m) + # σ0, ⌜n⌝ < ⌜n0⌝ + # σ0 ⊦ ((σ to n0 + m) + σ0)[m + n] = ((σ to m + n0) from m)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

Additional information

Used simplifier rules: get-append-ls: ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m];%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;