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Proof step 5 from proof javavalues-in (previous step)

∀ val. val ∈ mkjavavalues(refs0) ↔ is_referencevalue(val) ∧ val .refval ∈ refs0 ⊦ val ∈ mkjavavalues(r + refs0) ↔ is_referencevalue(val) ∧ val .refval ∈ r + refs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ mkjavavalues(r ' + refs) = refval(r) ' + mkjavavalues(refs)
Proof step 4:
∀ val. val ∈ mkjavavalues(refs0) ↔ is_referencevalue(val) ∧ val .refval ∈ refs0 ⊦ val = refval(r) ∨ val ∈ mkjavavalues(refs0) ↔ is_referencevalue(val) ∧ (val .refval = r ∨ val .refval ∈ refs0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ r ∈ r0 ' + refs ↔ r ≠ r0 ∧ ¬ r ∈ refs], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ val ∈ val0 ' ↔ val = val0], [CSIMPSEQ : ⊦ val ∈ vals + vals0 ↔ ¬ ¬ (val ∈ vals ∨ val ∈ vals0)], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]