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Proof step 3 from proof less-same (previous step)

n + 1 < #(a0 + x0), ∀ n, a. n + 1 < # x0 → less#(a, x0[n, x0[n + 1]][n + 1, x0[n]]) = less#(a, x0) ∨ x0 = [] ⊦ a0 + x0 = [], less#(a, (a0 + x0)[n, (a0 + x0)[n + 1]][n + 1, (a0 + x0)[n]]) = less#(a, a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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 = []

Rule argument

Additional information

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