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

m ≤ # x, n0 +1 ≤ # x, m ≤ n0 +1, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ x to n0 +1 to m = x to m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n0 < # x, m ≤ n0 + 1, m ≤ # x, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ x to n0 + 1 to m = x to m, x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)