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Proof step 59 from proof test (previous step)

buf1 = mk(mkarray(2), 2, 0, 0), ar0 = buf1 .data ∧ a2 = a ∧ i0 = 0 ⊦ ⟪buf1 := buf1 .data:= ar0[i→n(i0), a2] .len:= ((buf1 .data:= ar0[i→n(i0), a2]) .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 58:
buf1 = mk(mkarray(2), 2, 0, 0), ar0 = buf1 .data ∧ a2 = a ∧ i0 = 0, buf = buf1 .data:= ar0[i→n(i0), a2] .len:= ((buf1 .data:= ar0[i→n(i0), a2]) .len + 1) ⊦ ⟪push#(b; buf)⟫ ⟪pop#(; buf, a0) ; assert#(a0 = a) ; push#(c; buf) ; pop#(; buf, a0) ; assert#(a0 = b) ; pop#(; buf, a0) ; assert#(a0 = c)⟫ true

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]