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Proof step 51 from proof write-known-p (previous step)

H1[r, o1] = H0 ∪ H, ((r => o1 * P))(H0), P0(H), rely(me, O1, H1[r, o1], O2, H2), r ∈ H1, H0 ⊥ H, ∀ r. r ∈ H0 → O2(r) = me ⊦ ((r => H1[r] * P))(H0[r, H1[r]]) ∧ (r0 ∈ H0 → O2(r0) = me)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 52:
r ∈ H ⊦ ((r => o0 * P))(H[r, o1]) ↔ o0 = o1 ∧ ((r =>_ * P))(H)
Proof step 53:
H1[r, o1] = H0 ∪ H, ((r => o1 * P))(H0), P0(H), rely(me, O1, H1[r, o1], O2, H2), r ∈ H1, r ∈ H0, H0 ⊥ H, ∀ r. r ∈ H0 → O2(r) = me ⊦ (H1[r] = H1[r] ∧ ((r =>_ * P))(H0)) ∧ (r0 ∈ H0 → O2(r0) = me)

Rule argument

spec: maplet name: map-putex seq: r ∈ H ⊦ ((r => o0 * P))(H[r, o1]) ↔ o0 = o1 ∧ ((r =>_ * P))(H) subst: [o1, H, P, o0, r] → [H1[r], H0, P, H1[r], r] flags: <all paths>

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) Used terms: H1[r], H0, P, H1[r], r