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Proof step 97 from proof validrefs-subset (previous step)

∀ str, ty, st0, st, r. validrefs(str, acc, ty, st0) ∧ st ⊆ st0 ∧ ¬ junkref(str, r, st) ∧ acc ≠ [] ∧ acc .first = r → validrefs(str, acc, ty, st) ⊦ validrefs(str, r0 + acc, ty, st0) ∧ st ⊆ st0 ∧ ¬ junkref(str, r, st) ∧ r0 + acc ≠ [] ∧ (r0 + acc) .first = r → validrefs(str, r0 + acc, ty, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 96:
st ⊆ st0, validrefs(str, r0 ' + acc, ty, st0), ∀ str0, ty0, st1, st2. validrefs(str0, acc, ty0, st1) ∧ st2 ⊆ st1 ∧ ¬ junkref(str0, acc .first, st2) ∧ acc ≠ [] → validrefs(str0, acc, ty0, st2) ⊦ validrefs(str, r0 ' + acc, ty, st), junkref(str, r0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (r ' + refs) .first = r], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]