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Proof step 15 from proof nonnullrefs-in (previous step)

vals0[m - 1] = refval(r), m ≤ # vals0, ∀ r0. r0 ∈ vals0 .nonnullrefs ↔ r0 ≠ jvmref ∧ (∃ m. vals0[m] = refval(r0) ∧ m < # vals0) ⊦ r = r1, r1 = jvmref, r = jvmref, m = 0, r ∈ vals0 .nonnullrefs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
m = m0 + 1, vals0[m0] = refval(r), m ≤ # vals0, ∀ r0. r0 ∈ vals0 .nonnullrefs ↔ r0 ≠ jvmref ∧ (∃ m. vals0[m] = refval(r0) ∧ m < # vals0) ⊦ r = r1, r1 = jvmref, r = jvmref, m = 0, r ∈ vals0 .nonnullrefs

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, m0] 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, m0)]]