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Proof step 5 from proof eq (previous step)

a ' = ins≤(a, a0 + y0), ∀ a. a ' = ins≤(a, y0) → [] = y0 ⊦ [] = a0 + y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ' = ins≤(a, a0 ' + y0), (∃ a. a ' = ins≤(a, y0)) → [] = y0 ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)