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Proof step 2 from proof del1-02 (previous step)

∀ 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 3:
∀ a. ins≤(a, x0) -1l a = x0 ⊦ ins≤(a, a0 ' + x0) -1l a = a0 ' + x0

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list