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Proof step 3 from proof notin-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list