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Proof step 6 from proof extensionality-to (previous step)

# σ0 = # σ1, ⌜n⌝ < # σ1, ¬ # σ1 < ⌜n + 1⌝ → σ0 to n + 1 = σ1 to n + 1, ∀ n. ¬ # σ1 < ⌜n⌝ → σ0 to n = σ1 to n ⊦ σ0[n] = σ1[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: size-succ: ⊦ # σ < ⌜n⌝ + ⌜1⌝ ↔ ¬ ⌜n⌝ < # σ;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf