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Proof step 11 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) ⊦ a0 ' + x0 = [], less#(a, (a0 ' + x0)[n, x0[n]][n + 1, (a0 ' + x0).tail[n - 1]]) = (a0 < a ⊃ 1;0) + less#(a, x0), n = 0, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
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, x0[n - 1]]) = (a0 < a ⊃ 1;0) + less#(a, x0), n = 0, x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)