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Proof step 19 from proof pathex-put-array-badindex (previous step)

i < 0 ⊦ path∃(str, r, r1, st[r0 -- i ', val]) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r1 ∧ validrefs(str, refs, st[refs .first -- _type] .type, st))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st))
Proof step 18:
i < 0 ⊦ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r1 ∧ validrefs(str, refs, (st[r0 -- i ', val][r -- _type]) .type, st[r0 -- i ', val])) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r1 ∧ validrefs(str, refs, st[refs .first -- _type] .type, st))

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: pathex-def-simple antecedent: succedent: path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st)) substlist: [st, r0, r, str] → [st[r0 -- i ', val], r1, r, 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[r0 -- i ', val], r1, r, str)]]