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Proof step 3 from proof ex (previous step)

⊦ ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ [] = y + z

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], []

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a ';% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + [] = x; (s)% ⊦ a ' = b ' ↔ a = b; (s) Used terms: [], []