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Proof step 108 from proof leninv-succ (previous step)

m = m2 + 1, # nσ = ⌜m1⌝ + ⌜1⌝ , nσ[m0] = m2, stutf(nσ), nσ[m1] + 1 < m, m0 ≤ m1 ⊦ m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 109:
# nσ = ⌜m1⌝ + ⌜1⌝ , m0 ≤ m1, stutf(nσ), nσ[m1] < nσ[m0] ⊦ nσ[m0] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2