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Proof step 4 from proof validrefs-subset-equiv-01 (previous step)

¬ (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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
st ⊆ st0 ⊦ validrefs(str, r ' + refs, ty, st0) ↔ validrefs(str, r ' + refs, ty, st), junkref(str, r, st), validref(str, r, st[r -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]