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Proof step 6 from proof get-from-01 (previous step)

m1 + m0 < # x0, ∀ m0, x0, a. m1 + m0 ≤ # x0 ∧ (∀ m, m0. m + m0 < # x0 → (x0 from m)[m0] = x0[m + m0] ∨ x0 = []) → (a ' + x0 from m1)[m0] = (a ' + x0)[m1 + m0], ∀ m, m2. m + m2 < # x0 → (x0 from m)[m2] = x0[m + m2] ⊦ (x0 from m1)[m0] = x0[m1 + m0], x0 = []

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: m1, m0

Additional information

Used terms: m1, m0