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Proof step 5 from proof delpos-in (previous step)

n = n0 +1, a ∈ (a0 ' + x0) -1l n, ∀ a, n. a ∈ x0 -1l n → a ∈ x0 ⊦ a = a0, a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ x0 -1l n0, ∀ a, n. a ∈ x0 -1l n → a ∈ x0 ⊦ a = a0, a ∈ x0

Rule argument

Additional information

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