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Proof step 9 from proof eq (previous step)

a0 < a, ins≤(a, a0 ' + x0) = ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a0 ' + ins≤(a, x0) = ins≤(a, y), a0 < a, a0 ∈ ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x);%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list