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

# x ≤ pos(a, x), a ∈ x ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: x

Additional information

Used simplifier rules: pos-e: ⊦ pos(a, []) = 0;%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 [], +;;%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%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%size-01: ⊦ # [] = 0; from specification list-data%in-empty: ⊦ ¬ a ∈ []; from specification list