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Proof step 10 from proof nonnullrefs-dups (previous step)

r0 ∈ vals0 .nonnullrefs ∨ dups(vals0 .nonnullrefs), dups(vals0 .nonnullrefs) ↔ (∃ r, m, n. vals0[m] = refval(r) ∧ vals0[n] = refval(r) ∧ r ≠ jvmref ∧ m ≠ n ∧ m < # vals0 ∧ n < # vals0) ⊦ dups(vals0 .nonnullrefs), ∃ r, m, n. (refval(r0) ' + vals0)[m] = refval(r) ∧ (refval(r0) ' + vals0)[n] = refval(r) ∧ r ≠ jvmref ∧ m ≠ n ∧ ¬ # vals0 < m ∧ ¬ # vals0 < n, r0 = jvmref

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
r0 ∈ vals0 .nonnullrefs ⊦ r0 = jvmref, dups(vals0 .nonnullrefs), ∃ r, m, n. (refval(r0) ' + vals0)[m] = refval(r) ∧ (refval(r0) ' + vals0)[n] = refval(r) ∧ r ≠ jvmref ∧ m ≠ n ∧ ¬ # vals0 < m ∧ ¬ # vals0 < n, ∃ r, m, n. vals0[m] = refval(r) ∧ vals0[n] = refval(r) ∧ r ≠ jvmref ∧ m ≠ n ∧ ¬ # vals0 ≤ m ∧ ¬ # vals0 ≤ n

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : nat, , lels, ⊦ ¬ m < n ↔ n ≤ m])]]