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Proof step 42 from proof test (previous step)

buf = mk(mkarray(2)[0, a][1, b], 2, 0, 2), a2 = a1 ∧ ar0 = buf .data ∧ i0 = buf .first ⊦ ⟪buf0 := buf .first:= ((buf .first + 1) % buf .size)⟫ ⟪buf0 := buf0 .len:= (buf0 .len - 1)⟫ ⟪assert#(ar0[i→n(i0)] = a) ; push#(c; buf0) ; a2 := ar0[i→n(i0)] ; pop#(; buf0, a2) ; assert#(a2 = b) ; pop#(; buf0, a2) ; assert#(a2 = c)⟫ true

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 41:
buf = mk(mkarray(2)[0, a][1, b], 2, 0, 2), a2 = a1 ∧ ar0 = buf .data ∧ i0 = buf .first ⊦ ⟪buf0 := buf .first:= ((buf .first + 1) % buf .size) .len:= ((buf .first:= ((buf .first + 1) % buf .size)) .len - 1)⟫ ⟪assert#(ar0[i→n(i0)] = a) ; push#(c; buf0) ; a2 := ar0[i→n(i0)] ; pop#(; buf0, a2) ; assert#(a2 = b) ; pop#(; buf0, a2) ; assert#(a2 = c)⟫ true

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]