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Proof step 3 from proof pos-equal (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
pos(b, a0 ' + x0) = pos(a, x0) + 1, a ∈ x0, ∀ b, a. pos(b, x0) = pos(a, x0) ∧ a ∈ x0 → a = b ⊦ a = b, a = a0

Rule argument

Additional information

Used simplifier rules: pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1;%pos-y: ⊦ pos(a, a ' + x) = 0;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ a + x = a ' + x; from specification list