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Proof step 35 from proof nonnullrefs-dups (previous step)

(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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 34:
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

Rule argument

[x0lemmaarg: spec-name: nat instance-name: lemma-name: elim-pred-c antecedent: n ≠ 0 succedent: m = n - 1 ↔ n = m + 1 substlist: [n, m] → [m, m1] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (m, m1)]]