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Proof step 10 from proof sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a0 < b, ∀ a, b. ins≤(a, ins≤(b, x0)) = ins≤(b, ins≤(a, x0)) ⊦ ins≤(a, a0 ' + ins≤(b, x0)) = ins≤(b, ins≤(a, a0 ' + x0))

Rule argument

Additional information

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