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Proof step 34 from proof nonnullrefs-dups (previous step)

m = m1 + 1, (refval(r0) ' + vals0)[n] = refval(r), vals0[m1] = 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
vals0[m1] = refval(r), (refval(r0) ' + vals0)[n] = refval(r), m1 < # vals0, n ≤ # vals0 ⊦ r = jvmref, m1 + 1 = n, r0 = jvmref, vals0 = [], dups(vals0 .nonnullrefs), r0 ∈ 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])]]