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Proof step 7 from proof del1-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a0 < a, ∀ a. ins≤(a, x0) -1l a = x0 ⊦ ins≤(a, x0) -1l a = x0, a = a0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = x + z ↔ y = z; (s)