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Proof step 31 from proof validrefs-has-null (previous step)

∀ refs0, str, ty, st. jvmref ∈ acc ∧ refs0 ≠ [] → ¬ validrefs(str, acc + refs0, ty, st) ⊦ jvmref ∈ r + acc ∧ refs0 ≠ [] → ¬ validrefs(str, (r + acc) + refs0, ty, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ validrefs(str, jvmref ' + refs, ty, st) ↔ refs = []
Proof step 30:
validrefs(str, (r ' + acc) + refs0, ty, st), jvmref ∈ acc, ∀ refs, str0, ty0, st0. refs ≠ [] → ¬ validrefs(str0, acc + refs, ty0, st0) ⊦ refs0 = [], jvmref = r

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]