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Proof step 5 from proof del1-01-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 < a, ordered≤(a0 ' + x0), ∀ a1, b0. ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ∨ a1 = b0 ⊦ a = b, ins≤(a, (a0 ' + x0) -1l b) = (a0 ' + ins≤(a, x0)) -1l b

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