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Proof step 6 from proof ins-less-last (previous step)

a1 < a, ∀ a, a0. ins≤(a, x0 + a0 ') = ins≤(a, x0) + a0 ' ∨ a0 < a ⊦ ins≤(a, (a1 ' + x0) + a0 ') = ins≤(a, a1 ' + x0) + a0 ', a0 < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a1 < a, ∀ a, a0. ins≤(a, x0 + a0 ') = ins≤(a, x0) + a0 ' ∨ a0 < a ⊦ ins≤(a, x0 + a0 ') = ins≤(a, x0) + a0 ', a0 < a

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: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)