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Proof step 6 from proof less-same (previous step)

n < # x0, ∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ less#(a, (a0 ' + x0)[n, x0[n]][n + 1, (a0 ' + x0)[n]]) = (a0 < a ⊃ 1;0) + less#(a, x0), x0 = [], a0 ' + x0 ≠ [] ∧ n ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ less#(a, (x0.head ' + x0)[1, a0]) = (a0 < a ⊃ 1;0) + less#(a, x0), x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ (a ' + x)[0] = a; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ (b ' + x)[0, a] = a + x; (s)% ⊦ x ≠ [] → x[0] = x.head; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)