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Proof step 24 from proof nodes-cons-rec (previous step)

# ax = # ndx, gstore[@(m3, m4, m5)] = mkgnode(nd0, m6, boolvar1, boolvar2), gstore[@(m, m0, m1)] = mkgnode(nd, m2, boolvar, boolvar0), ax[n - 1] = @(m3, m4, m5), @(m, m0, m1) ∈ gstore, n ≤ # ndx, ∀ n. n < # ndx → ax[n] ∈ gstore ∧ ¬ ax[n].lnum ∈ ginval ∧ gstore[ax[n]].nd = ndx[n] ⊦ n = 0, m ∈ ginval, @(m3, m4, m5) ∈ gstore ∧ ¬ m3 ∈ ginval ∧ nd0 = (nd ' + ndx)[n]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
# ax = # ndx, ax[m7] = @(m3, m4, m5), gstore[@(m, m0, m1)] = mkgnode(nd, m2, boolvar, boolvar0), gstore[@(m3, m4, m5)] = mkgnode(nd0, m6, boolvar1, boolvar2), n = m7 + 1, @(m, m0, m1) ∈ gstore, n ≤ # ndx, ∀ n. n < # ndx → ax[n] ∈ gstore ∧ ¬ ax[n].lnum ∈ ginval ∧ gstore[ax[n]].nd = ndx[n] ⊦ n = 0, m ∈ ginval, @(m3, m4, m5) ∈ gstore ∧ ¬ m3 ∈ ginval ∧ nd0 = (nd ' + ndx)[n]

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m7]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m7