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Proof step 15 from proof disjoint-symmetric (conclusion)

⊦ disjoint(str, r ' + r0 ', st) ↔ disjoint(str, r0 ' + r ', st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ disjoint(str, refs, st) ↔ ¬ (∃ refs1, refs2. refs1 ≠ [] ∧ refs2 ≠ [] ∧ refs1 .first ∈ refs ∧ refs2 .first ∈ refs ∧ refs1 .first ≠ refs2 .first ∧ refs1 .last = refs2 .last ∧ refs1 .last ≠ jvmref ∧ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ∧ validrefs(str, refs2, st[refs2 .first -- _type] .type, st))
Proof step 14:
⊦ ¬ (∃ refs1, refs2. refs1 ≠ [] ∧ refs2 ≠ [] ∧ refs1 .first ∈ r ' + r0 ' ∧ refs2 .first ∈ r ' + r0 ' ∧ refs1 .first ≠ refs2 .first ∧ refs1 .last = refs2 .last ∧ refs1 .last ≠ jvmref ∧ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ∧ validrefs(str, refs2, st[refs2 .first -- _type] .type, st)) ↔ disjoint(str, r0 ' + r ', st)

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: disjoint-def antecedent: succedent: disjoint(str, refs, st) ↔ ¬ (∃ refs1, refs2. refs1 ≠ [] ∧ refs2 ≠ [] ∧ refs1 .first ∈ refs ∧ refs2 .first ∈ refs ∧ refs1 .first ≠ refs2 .first ∧ refs1 .last = refs2 .last ∧ refs1 .last ≠ jvmref ∧ validrefs (str, refs1, st[refs1 .first -- _type] .type, st) ∧ validrefs (str, refs2, st[refs2 .first -- _type] .type, st)) substlist: [st, refs, str] → [st, r ' + r0 ', str] add preconditions?: F is rewrite lemma?: T current spec?: T rotate lemma?: F replace all?: T paths: ((1, 0, 0, 1, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r ' + r0 ', str)]]