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Proof step 1 from proof lr-04 (conclusion)

⊦ # x.rest < # x ↔ x ≠ []

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: size-01: ⊦ # [] = 0;%length-rest: ⊦ x ≠ [] → # x.rest = (# x)-1;%l: ⊦ # x = 0 ↔ x = [];%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%pred-03: ⊦ n -1 = n - 1; from specification nat%zero-01: ⊦ ¬ m < 0; from specification nat-basic1