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Proof step 9 from proof del1-01-02 (previous step)

a0 < a, ordered≤(a0 ' + x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = b, b = a0, ins≤(a, a0 ' + (x0 -1l b)) = (a0 ' + ins≤(a, x0)) -1l b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a0 < a, ordered≤(a0 ' + x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = b, b = a0, ins≤(a, x0 -1l b) = ins≤(a, x0) -1l b

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x);% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)