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Proof step 4 from proof del1-02 (previous step)

a0 < a, ∀ a. ins≤(a, x0) -1l a = x0 ⊦ ins≤(a, a0 ' + x0) -1l a = a0 ' + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 < a, ∀ a. ins≤(a, x0) -1l a = x0 ⊦ (a0 ' + ins≤(a, x0)) -1l 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