[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 26 from proof mapex-new (previous step)

H ++ r = (∅ ++ r)[r, o] ∪ H0, P(H0) ⊦ H = H0, r = null, P(H), r ∈ H0, r ∈ H

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 28:
H ++ r = (∅ ++ r)[r, o] ∪ H0, P(H0) ⊦ r = null, P(H), r ∈ H, r ∈ H0, ∀ r. (r ∈ H0 ↔ r ∈ H) ∧ (r ∈ H0 → H0[r] = H[r])

Rule argument

spec: heap name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [H, H0] flags: <all paths>

Additional information

Used terms: H, H0