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Proof step 2 from proof l-01 (previous step)

# x = # x - 1, a ∈ x ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: x

Additional information

Used simplifier rules: list freely generated by [], +;;%c: ⊦ a + x = a ' + x; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%size-01: ⊦ # [] = 0; from specification list-data%in-empty: ⊦ ¬ a ∈ []; from specification list