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Proof step 3 from proof inv-add (previous step)

# xs = xs_len, # xs < h, # ys = 0 ∨ # ys = h ⊦ ⟪if xs_len + 1 = h then {ys := a + xs; xs_len := 0; xs := []} else {xs := a + xs; xs_len := 1 + xs_len}⟫ (# xs < h ∧ # xs = xs_len ∧ (# ys = 0 ∨ # ys = h))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 4:
# xs = xs_len, xs_len + 1 = h, # xs < h, # ys = 0 ∨ # ys = h ⊦ ⟪ys := a + xs; xs_len := 0; xs := []⟫ (# xs < h ∧ # xs = xs_len ∧ (# ys = 0 ∨ # ys = h))
Proof step 7:
# xs = xs_len, # xs < h, # ys = 0 ∨ # ys = h ⊦ ⟪xs := a + xs; xs_len := 1 + xs_len⟫ (# xs < h ∧ # xs = xs_len ∧ (# ys = 0 ∨ # ys = h)), xs_len + 1 = h

Rule argument

pos: right 1

Proof step 3 from proof inv-add (previous step)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

Additional information