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Proof step 18 from proof nonnullrefs-dups (previous step)

n = m + 1, vals0[m] = refval(r), n ≤ # vals0 ⊦ r = jvmref, n = 0, dups(vals0 .nonnullrefs), r ∈ vals0 .nonnullrefs, ∃ r1, m0, n0. vals0[m0] = refval(r1) ∧ vals0[n0] = refval(r1) ∧ r1 ≠ jvmref ∧ m0 ≠ n0 ∧ ¬ # vals0 ≤ m0 ∧ ¬ # vals0 ≤ n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
vals0[m] = refval(r), m < # vals0 ⊦ r = jvmref, vals0 = [], r ∈ vals0 .nonnullrefs, dups(vals0 .nonnullrefs), ∃ r1, m0, n0. vals0[m0] = refval(r1) ∧ vals0[n0] = refval(r1) ∧ r1 ≠ jvmref ∧ m0 ≠ n0 ∧ ¬ # vals0 ≤ m0 ∧ ¬ # vals0 ≤ n0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list-data, javavalues, l, ⊦ # vals = 0 ↔ vals = []], [CSIMPNAMEDFORWARD : nat-basic1, , f, ⊦ m < n → n ≠ 0], [CSIMPNAMEDSEQ : nat, , succ-01, ⊦ m + 1 ≤ n ↔ m < n], [CSIMPNAMEDSEQ : nat-basic2, , c, ⊦ m + n = n + m], [CSIMPNAMEDSEQ : nat-basic2, , a, ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPNAMEDSEQ : nat-basic2, , oz, ⊦ m + 1 ≠ 0])]]