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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 [], +;;% ⊦ a + x = a ' + x; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ # a ' = 0 +1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s)% ⊦ # [] = 0; (s)% ⊦ ¬ a ∈ []; (s)