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Proof step 12 from proof ex (previous step)

ins≤(a, y + z) = y + a ' + z, ∀ a. ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ y + z = y0 + z0 ⊦ a ' + a0 ' + y = a0 ' + y + a ', a0 < a, ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ a0 ' + y + z = y0 + z0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: [], a0 ' + y + z

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;%Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list Used terms: [], a0 ' + y + z