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Proof step 7 from proof completeness (previous step)

x = [], x2 = x ∧ d0 = 0 ⊦ 〈if is_empty(x2) then to0 := fail else let h = head(x2) in if h < d0 then to0 := fail else if h = d0 then { x2 := pop(x2) ; to0 := some(Leaf) } else let to2 = fail, to1 = fail in { build_rec#(d0 + 1; x2, to2) ; if to2 = fail then to0 := fail else { build_rec#(d0 + 1; x2, to1) ; if to1 = fail then to0 := fail else to0 := some(Node(to2 .t, to1 .t)) } }〉〈to1 := to0, x1 := x2 ; if x1 ≠ [] then to1 := fail〉 true

Heuristic symbolic execution applied rule if positive right

And obtained the following new goals

Proof step 6:
x = [], x2 = x ∧ d0 = 0 ⊦ 〈to0 := fail〉〈to1 := to0, x1 := x2 ; if x1 ≠ [] then to1 := fail〉 true

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ is_empty(x) ↔ x = []])]]