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Proof step 16 from proof less-same (previous step)

m + 1 < # x0, ∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ a0 ' + x0[m, x0[m + 1]] = [], 2 + m = 0, less#(a, (a0 ' + x0[m, x0[m + 1]]).head ' + (a0 ' + x0[m, x0[m + 1]]).tail[(2 + m)-1, x0[m]]) = (a0 < a ⊃ 1;0) + less#(a, x0), x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
m + 1 < # x0, ∀ n0, a1. n0 + 1 < # x0 → less#(a1, x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = less#(a1, x0) ⊦ less#(a, x0[m, x0[m + 1]][1 + m, x0[m]]) = less#(a, x0), x0 = []

Rule argument

Additional information

Used simplifier rules: less-rec: ⊦ less#(a, b ' + x) = (b < a ⊃ 1;0) + less#(a, x);% ⊦ a ' + x ≠ []; (s)%n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n -1 = n - 1; (s)% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).head = a; (s)