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Proof step 3 from proof possize-notin (previous step)

pos(a, a0 + x0) = #(a0 + x0), a ∈ a0 + x0, ∀ a. ¬(pos(a, x0) = # x0 ∧ a ∈ x0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
pos(a, x0) = # x0, a ∈ x0, ∀ a1. pos(a1, x0) ≠ pos(a, x0) ∨ ¬ a1 ∈ x0 ⊦ a = a0

Rule argument

Additional information

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