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Proof step 12 from proof harness-2 (previous step)

x = 2 ' + 2 ', to6 = fail, x2 = x ∧ d0 = 3, h0 = head(x2), to2 = fail ∧ x0 = x2 ⊦ 〈if fail = fail then to0 := fail else to0 := some(Node(Leaf, fail .t))〉〈to1 := to0 ; if x0 ≠ [] then to1 := fail〉 to1 = fail

Heuristic symbolic execution applied rule if positive right

And obtained the following new goals

Proof step 11:
x = 2 ' + 2 ', to6 = fail, x2 = x ∧ d0 = 3, h0 = head(x2), to2 = fail ∧ x0 = x2 ⊦ 〈to0 := fail〉〈to1 := to0 ; if x0 ≠ [] then to1 := fail〉 to1 = fail

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ()]]