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Proof step 6 from proof pos-equal (previous step)

b ≠ a0, pos(b, x0) + 1 = pos(a, x0) + 1, a ∈ x0, ∀ b, a. pos(b, x0) = pos(a, x0) ∧ a ∈ x0 → a = b ⊦ a = b, a = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
pos(b, x0) = pos(a, x0), a ∈ x0, ∀ b, a. pos(b, x0) = pos(a, x0) ∧ a ∈ x0 → a = b ⊦ a = b, a = a0, b = a0

Rule argument

Additional information

Used simplifier rules: 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