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Proof step 12 from proof split-first (previous step)

⊦ ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = # y + 1

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a ' + y, z

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%len-05: ⊦ #(a ' + x) = (# x)+1;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: a ' + y, z