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Proof step 59 from proof junkrefs-notin (previous step)

∀ str, r, refs0, ty, st. (junkrefs(str, acc + r ' + refs0, ty, st) ↔ junkrefs(str, acc + r ', ty, st)) ∨ r ∈ st ⊦ junkrefs(str, (r0 + acc) + r ' + refs0, ty, st) ↔ junkrefs(str, (r0 + acc) + r ', ty, st), r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
∀ str, r, refs0, ty, st. (junkrefs(str, acc + r ' + refs0, ty, st) ↔ junkrefs(str, acc + r ', ty, st)) ∨ r ∈ st ⊦ junkrefs(str, (r0 ' + acc) + r ' + refs0, ty, st) ↔ junkrefs(str, (r0 ' + acc) + r ', ty, st), r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r + refs = r ' + refs])]]