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Proof step 5 from proof validrefs-subset-equiv-01 (previous step)

st ⊆ st0 ⊦ validrefs(str, r ' + refs, ty, st0) ↔ validrefs(str, r ' + refs, ty, st), junkref(str, r, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ validref(str, r, st[r -- _type] .type, st) ∧ st ⊆ st0 → (validrefs(str, r ' + refs, ty, st0) ↔ validrefs(str, r ' + refs, ty, st))
Proof step 4:
¬ (validref(str, r, st[r -- _type] .type, st) ∧ st ⊆ st0), st ⊆ st0 ⊦ validrefs(str, r ' + refs, ty, st0) ↔ validrefs(str, r ' + refs, ty, st), junkref(str, r, st)

Rule argument

[crewrite-arg: validrefs-subset-equiv , F , [str, r, st, st0, refs, ty] → [str, r, st, st0, refs, ty] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (str, r, st, st0, refs, ty)]]