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Proof step 4 from proof get (previous step)

a ∈ a0 + x0, ∀ a. a ∈ x0 → x0[pos(a, x0)] = a ⊦ (a0 + x0)[pos(a, a0 + x0)] = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ x0, ∀ a. a ∈ x0 → x0[pos(a, x0)] = a ⊦ x0[pos(a, x0)] = a, a = a0

Rule argument

Additional information

Used simplifier rules: pos-y: ⊦ pos(a, a ' + x) = 0;%get-zero-01: ⊦ (a ' + x)[0] = a;%pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1;%get-succ: ⊦ (a ' + x)[n + 1] = x[n];%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)