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Proof step 3 from proof inl-01 (previous step)

a ∈ x, # x ≤ pos(a, x) ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: x

Additional information

Used simplifier rules: pos-y: ⊦ pos(a, a ' + x) = 0;%pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1;%inl: ⊦ a ∈ x → pos(a, x) < # x;%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% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ ¬ n < 0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ a ∈ []; (s)