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Proof step 4 from proof alloc (
previous step
)
P(H) ⊦ ∃ r0. ¬ r0 ∈ H ∧ r0 ≠ null
User applied rule apply lemma
And obtained the following new goals
Proof step 5
:
⊦ ∃ r. ¬ r ∈ H ∧ r ≠ null
Proof step 6
:
P(H), ∃ r. ¬ r ∈ H ∧ r ≠ null ⊦ ∃ r0. ¬ r0 ∈ H ∧ r0 ≠ null
Rule argument
spec: ops-impl name: infinite-mem seq: ⊦ ∃ r. ¬ r ∈ H ∧ r ≠ null subst: [H] → [H]
Additional information
Used terms: H