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Proof step 74 from proof validrefs-subset-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
validrefs(str, r0 ' + acc, ty, st), st ⊆ st0, ∀ str0, ty0, st1, st2. validrefs(str0, acc, ty0, st1) ∧ st1 ⊆ st2 ∧ ¬ junkref(str0, acc .first, st1) ∧ acc ≠ [] → validrefs(str0, acc, ty0, st2) ⊦ r0 = jvmref, junkref(str, r0, st), validrefs(str, r0 ' + acc, ty, st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])]]