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Proof step 67 from proof harness-1 (previous step)

x = 3 ' + 3 ' + 2 ', to4 = fail ⊦ 〈x0 := x, d := 1〉〈if is_empty(x0) then to := fail else let h = head(x0) in if h < d then to := fail else if h = d then { x0 := pop(x0) ; to := some(Leaf) } else let to0 = fail, to1 = fail in { build_rec#(d + 1; x0, to0) ; if to0 = fail then to := fail else { build_rec#(d + 1; x0, to1) ; if to1 = fail then to := fail else to := some(Node(to0 .t, to1 .t)) } }〉〈to4 := to ; if to4 = fail then to0 := fail else to0 := some(Node(Leaf, to4 .t))〉〈to1 := to0 ; if x0 ≠ [] then to1 := fail〉 to1 = some(Node(Leaf, Node(Node(Leaf, Leaf), Leaf)))

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 66:
x = 3 ' + 3 ' + 2 ', to4 = fail, x1 = x ∧ d0 = 1 ⊦ 〈if is_empty(x1) then to := fail else let h = head(x1) in if h < d0 then to := fail else if h = d0 then { x1 := pop(x1) ; to := some(Leaf) } else let to0 = fail, to1 = fail in { build_rec#(d0 + 1; x1, to0) ; if to0 = fail then to := fail else { build_rec#(d0 + 1; x1, to1) ; if to1 = fail then to := fail else to := some(Node(to0 .t, to1 .t)) } }〉〈to4 := to ; if to4 = fail then to0 := fail else to0 := some(Node(Leaf, to4 .t))〉〈to1 := to0 ; if x1 ≠ [] then to1 := fail〉 to1 = some(Node(Leaf, Node(Node(Leaf, Leaf), Leaf)))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]