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Proof step 36 from proof nonnullrefs-dups (previous step)

m ≠ 0, vals0[m -1] = refval(r), (refval(r0) ' + vals0)[n] = refval(r), n ≤ # vals0, m ≤ # vals0 ⊦ r0 = jvmref, m = n, r = jvmref, r0 ∈ vals0 .nonnullrefs, dups(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 35:
(refval(r0) ' + vals0)[n] = refval(r), vals0[m - 1] = refval(r), n ≤ # vals0, m ≤ # vals0 ⊦ r0 = jvmref, m = n, r = jvmref, m = 0, r0 ∈ 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 : nat, , pred-03, ⊦ n -1 = n - 1])]]