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Proof step 7 from proof from-append-tofrom (previous step)

¬ ¬ #(σ from n) < ⌜n1⌝ ⊦ # σ < ⌜n⌝ + ⌜n1⌝ , #((σ from n) to n1) = ⌜n1⌝ ∧ (⌜n0⌝ < ⌜n⌝ + #((σ from n) to n1) → ((σ to n) + ((σ from n) to n1))[n0] = (σ to n + n1)[n0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
#(σ from n) < ⌜n1⌝ ⊦ # σ < ⌜n⌝ + ⌜n1⌝ , #((σ from n) to n1) = ⌜n1⌝ ∧ (⌜n0⌝ < ⌜n⌝ + #((σ from n) to n1) → ((σ to n) + ((σ from n) to n1))[n0] = (σ to n + n1)[n0])

Proof step 7 from proof from-append-tofrom (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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