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Proof step 12 from proof harness-1 (previous step)

to3 = fail, x0 = 2 ', x1 = x0 ∧ d0 = 2, h0 = head(x1) ⊦ 〈to0 := some(Node(Leaf, some(Node(Node(Leaf, Leaf), some(Leaf) .t)) .t))〉〈to1 := to0 ; if pop(x1) ≠ [] 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 11:
to3 = fail, x0 = 2 ', x1 = x0 ∧ d0 = 2, h0 = head(x1) ⊦ 〈to1 := some(Node(Leaf, some(Node(Node(Leaf, Leaf), some(Leaf) .t)) .t))〉〈if pop(x1) ≠ [] then to1 := fail〉 to1 = some(Node(Leaf, Node(Node(Leaf, Leaf), Leaf)))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]