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Proof step 62 from proof test (previous step)

buf1 = mk(mkarray(2), 2, 0, 0) ⊦ ⟪ar := buf1 .data, a1 := a, i := 0⟫ ⟪if 0 ≤ i ∧ i < n→i(# ar) then ar := ar[i→n(i), a1] else abort⟫ ⟪buf1 := buf1 .data:= ar .len:= ((buf1 .data:= ar) .len + 1)⟫ ⟪push#(b; buf1) ; pop#(; buf1, a0) ; assert#(a0 = a) ; push#(c; buf1) ; pop#(; buf1, a0) ; assert#(a0 = b) ; pop#(; buf1, a0) ; assert#(a0 = c)⟫ true

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 61:
buf1 = mk(mkarray(2), 2, 0, 0), ar0 = buf1 .data ∧ a2 = a ∧ i0 = 0 ⊦ ⟪if 0 ≤ i0 ∧ i0 < n→i(# ar0) then ar0 := ar0[i→n(i0), a2] else abort⟫ ⟪buf1 := buf1 .data:= ar0 .len:= ((buf1 .data:= ar0) .len + 1)⟫ ⟪push#(b; buf1) ; pop#(; buf1, a0) ; assert#(a0 = a) ; push#(c; buf1) ; pop#(; buf1, a0) ; assert#(a0 = b) ; pop#(; buf1, a0) ; assert#(a0 = c)⟫ true

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]