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Proof step 16 from proof nonnullrefs-in-key (previous step)

r ∈ fis0 .vals .nonnullrefs → fis0[sk] = refval(r) ∧ sk ∈ fis0 ⊦ r = jvmref, r0 = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × refval(r0)) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0), r ≠ r0 ∧ ¬ r ∈ fis0 .vals .nonnullrefs

User applied rule cut formula

And obtained the following new goals

Proof step 8:
r = r0, r ∈ fis0 .vals .nonnullrefs → fis0[sk] = refval(r) ∧ sk ∈ fis0 ⊦ r = jvmref, r0 = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × refval(r0)) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0), r ≠ r0 ∧ ¬ r ∈ fis0 .vals .nonnullrefs
Proof step 15:
r ∈ fis0 .vals .nonnullrefs → fis0[sk] = refval(r) ∧ sk ∈ fis0 ⊦ r = r0, r = jvmref, r0 = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × refval(r0)) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0), r ≠ r0 ∧ ¬ r ∈ fis0 .vals .nonnullrefs

Rule argument

[fma-arg: r = r0]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (r = r0)]]