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Proof step 4 from proof in-01-02 (previous step)

∀ 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 5:
∀ a, b. (a ∈ ins≤(b, x0) ↔ a ∈ x0) ∨ a = b ⊦ a ∈ ins≤(b, a0 ' + x0) ↔ a ∈ a0 ' + x0, a = b

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)