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Proof step 19 from proof disjoint-subset (previous step)

st ⊆ st1, disjoint(str, refs, st), ∀ r. r ∈ refs → ¬ junkref(str, r, st) ⊦ disjoint(str, refs, st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ disjoint(str, refs, st) ↔ ¬ (∃ r, r0, r1. r ∈ refs ∧ r0 ∈ refs ∧ r ≠ r0 ∧ r1 ≠ jvmref ∧ path∃(str, r, r1, st) ∧ path∃(str, r0, r1, st))
Proof step 18:
st ⊆ st1, disjoint(str, refs, st), ∀ r. r ∈ refs → ¬ junkref(str, r, st) ⊦ ∀ r, r0, r1. ¬ (r ∈ refs ∧ r0 ∈ refs ∧ r ≠ r0 ∧ r1 ≠ jvmref ∧ path∃(str, r, r1, st1) ∧ path∃(str, r0, r1, st1))

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: disjoint-def-pathex antecedent: succedent: disjoint(str, refs, st) ↔ ¬ (∃ r, r0, r1. r ∈ refs ∧ r0 ∈ refs ∧ r ≠ r0 ∧ r1 ≠ jvmref ∧ path∃(str, r, r1, st) ∧ path∃(str, r0, r1, st)) substlist: [st, refs, str] → [st1, refs, str] add preconditions?: F is rewrite lemma?: T current spec?: T rotate lemma?: F replace all?: T paths: ((1, 0, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st1, refs, str)]]