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Proof step 4 from proof in-04 (previous step)

a0 < a, ∀ a. a ∈ ins≤(a, x0) ⊦ a ∈ ins≤(a, a0 ' + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 < a, ∀ a. a ∈ ins≤(a, x0) ⊦ a = a0, a ∈ ins≤(a, x0)

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x);%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list