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Proof step 22 from proof mapex-new (previous step)

((r => o * P))(H ++ r) ⊦ r = null, P(H), r ∈ H

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 24:
∃ H0, H1. H0 ⊥ H1 ∧ H ++ r = H0 ∪ H1 ∧ ((r => o))(H0) ∧ P(H1) ⊦ r = null, P(H), r ∈ H

Rule argument

spec: heap-sep name: sep seq: ⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)) subst: [H, P1, P0] → [H ++ r, P, r => o] flags: <all paths>

Additional information

Used terms: H ++ r, P, r => o