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

refval(r1) ' + vals0 ≠ [] ∧ m ≠ 0, ((refval(r1) ' + vals0) .rest)[m - 1] = refval(r), m ≤ # vals0, ∀ r0. r0 ∈ vals0 .nonnullrefs ↔ r0 ≠ jvmref ∧ (∃ m. vals0[m] = refval(r0) ∧ m < # vals0) ⊦ r = r1, r1 = jvmref, r = jvmref, r ∈ vals0 .nonnullrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
vals0[m - 1] = refval(r), m ≤ # vals0, ∀ r0. r0 ∈ vals0 .nonnullrefs ↔ r0 ≠ jvmref ∧ (∃ m. vals0[m] = refval(r0) ∧ m < # vals0) ⊦ r = r1, r1 = jvmref, r = jvmref, m = 0, r ∈ vals0 .nonnullrefs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, javavalues, e-01, ⊦ val ' + vals ≠ []], [CSIMPNAMEDSEQ : list, javavalues, a, ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPNAMEDSEQ : list, javavalues, c-07, ⊦ (val ' + vals) .rest = vals])]]