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Proof step 26 from proof harness-1 (previous step)

x = 3 ' + 2 ', to6 = fail, x2 = x ∧ d0 = 3, h0 = head(x2), to2 = some(Node(Leaf, some(Leaf) .t)) ∧ x0 = pop(x2), to3 = fail ⊦ 〈build_rec#(2; x0, to3)〉〈if to3 = fail then to := fail else to := some(Node(to2 .t, to3 .t))〉〈to5 := to ; if to5 = fail then to0 := fail else to0 := some(Node(Leaf, to5 .t))〉〈to1 := to0 ; if x0 ≠ [] then to1 := fail〉 to1 = some(Node(Leaf, Node(Node(Leaf, Leaf), Leaf)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
to3 = fail, x0 = 2 ' ⊦ 〈build_rec#(2; x0, to3)〉〈if to3 = fail then to := fail else to := some(Node(Node(Leaf, Leaf), to3 .t))〉〈to5 := to ; if to5 = fail then to0 := fail else to0 := some(Node(Leaf, to5 .t))〉〈to1 := to0 ; if x0 ≠ [] then to1 := fail〉 to1 = some(Node(Leaf, Node(Node(Leaf, Leaf), Leaf)))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ some(t) .t = t], [CSIMPSEQ : ⊦ (x + y) + z = x + y + z], [CSIMPSEQ : ⊦ pop(d ' + x) = x])]]