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Proof step 5 from proof ins-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 < a, ∀ a, y. sort(ins≤(a, x0) + y) = ins≤(a, sort(x0 + y)) ⊦ ins≤(a0, sort(ins≤(a, x0) + y)) = ins≤(a, ins≤(a0, sort(x0 + y)))

Rule argument

Additional information

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