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Proof step 31 from proof nonnullrefs-in (previous step)

val ' + vals0 ≠ [] ∧ m ≠ 0, ((val ' + vals0) .rest)[m - 1] = refval(r), m ≤ # vals0, ¬ is_referencevalue(val) ∨ val .refval = jvmref ⊦ r = jvmref, r ∈ vals0 .nonnullrefs, ∃ m0. vals0[m0] = refval(r) ∧ ¬ # vals0 ≤ m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
vals0[m - 1] = refval(r), m ≤ # vals0, ¬ is_referencevalue(val) ∨ val .refval = jvmref ⊦ r = jvmref, m = 0, r ∈ vals0 .nonnullrefs, ∃ m0. vals0[m0] = refval(r) ∧ ¬ # vals0 ≤ m0

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])]]