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Proof step 6 from proof in-01-02 (previous step)

a0 < b, ∀ a, b. (a ∈ ins≤(b, x0) ↔ a ∈ x0) ∨ a = b ⊦ a ∈ ins≤(b, a0 ' + x0) ↔ a ∈ a0 ' + x0, a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a0 < b, ∀ a, b. (a ∈ ins≤(b, x0) ↔ a ∈ x0) ∨ a = b ⊦ a ∈ ins≤(b, x0) ↔ a ∈ x0, a = b, a = a0

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x);% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)