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Proof step 6 from proof completeness-rec (previous step)

∀ j0, i0, x, x0, to. i→n(j0 - i0) < i→n(i1) → (x = x0 ∧ x0 ≠ [] ∧ i0 ≤ j0 ∧ max(x0) ≤ j0 → 〈build_rec#(i0; x, to)〉 (to = fail → (∀ t, x1. x0 ≠ depths(t, i0) + x1))), x = x0, x0 ≠ [], max(x0) ≤ i1 + i, 0 ≤ i1, h0 = head(x), ¬ h0 < i, h0 = i ⊦ 〈x := pop(x)〉〈to0 := some(Leaf)〉 (to0 = fail → (∀ t, x1. x0 ≠ depths(t, i) + x1))

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 5:
∀ j0, i0, x, x0, to. i→n(j0 - i0) < i→n(i1) → (x = x0 ∧ x0 ≠ [] ∧ i0 ≤ j0 ∧ max(x0) ≤ j0 → 〈build_rec#(i0; x, to)〉 (to = fail → (∀ t, x1. x0 ≠ depths(t, i0) + x1))), x = x0, x0 ≠ [], max(x0) ≤ i1 + i, 0 ≤ i1, h0 = head(x), ¬ h0 < i, h0 = i ⊦ 〈to0 := some(Leaf)〉 (to0 = fail → (∀ t, x1. x0 ≠ depths(t, i) + x1))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]