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Proof step 8 from proof to-to-leq-inf (previous step)

¬ m ≤ #(nf to n0), m ≤ n0 + 1, ∀ m, nf. m ≤ n0 → (nf to n0) to m = nf to m ⊦ (nf to n0) + nf(n0) ' to m = nf to m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ m, nf. m ≤ n0 → (nf to n0) to m = nf to m ⊦ (nf to n0) + nf(n0) ' to n0 + 1 = nf to n0 + 1

Rule argument

Additional information

Used simplifier rules: size-inf-to: ⊦ #(nf to n) = n;%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%les: ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat