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Proof step 2 from proof write-known (previous step)

o = o1, ((r => o0 * P))(H) ⊦ ⟪if ¬ r ∈ H then abort else H[r] := o⟫ ((r => o1 * P))(H)

Heuristic symbolic execution applied rule if negative right

And obtained the following new goals

Proof step 3:
o = o1, ((r => o0 * P))(H) ⊦ ⟪H[r] := o⟫ ((r => o1 * P))(H)

Rule argument

pos: right 1

Additional information

Used simplifier rules: sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep% ⊦ ((r => o * P))(H) → r ∈ H; (s)