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Proof step 51 from proof test (previous step)

buf1 = mk(mkarray(2)[0, a], 2, 0, 1), ar0 = buf1 .data ∧ a2 = b ∧ i0 = 1 ⊦ ⟪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)⟫ ⟪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 if positive right

And obtained the following new goals

Proof step 50:
buf1 = mk(mkarray(2)[0, a], 2, 0, 1), ar0 = buf1 .data ∧ a2 = b ∧ i0 = 1 ⊦ ⟪ar0 := ar0[i→n(i0), a2]⟫ ⟪buf1 := buf1 .data:= ar0 .len:= ((buf1 .data:= ar0) .len + 1)⟫ ⟪pop#(; buf1, a0) ; assert#(a0 = a) ; push#(c; buf1) ; pop#(; buf1, a0) ; assert#(a0 = b) ; pop#(; buf1, a0) ; assert#(a0 = c)⟫ true

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ mk(ar, i, i0, i1) .data = ar], [CSIMPSEQ : ⊦ # (ar[n, a]) = # ar], [CSIMPSEQ : ⊦ # mkarray(n) = n])]]