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Proof step 2 from proof le-one-01 (previous step)

# x + # y ≤ 1 ⊦ x = [], x.rest = [] ∧ y = []

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: x

Additional information

Used simplifier rules: gen: ⊦ ;%disj-1: ⊦ [] ≠ a + x;%size-02: ⊦ #(a + x) = (# x)+1;%l: ⊦ # x = 0 ↔ x = [];%sel-2: ⊦ (a + x).rest = x;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2