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Proof step 2 from proof psh-04 (previous step)

⊦ m < n -1 ↔ m +1 < n, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < n -1 ⊦ m + 1 = n, n = 0, n = 1, m ≠ n ∧ ¬ n < m

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%psh-01: ⊦ m < n0 → (n < n0 -1 ↔ n +1 < n0);%ls-04: ⊦ m < n ↔ ¬(m = n ∨ n < m);%s-04: ⊦ m < n + 1 ↔ ¬ n < m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%f: ⊦ m < n → n ≠ 0;%psh-03: ⊦ m ≠ 0 → (m -1 = n ↔ m = n +1);%one-06: ⊦ 1 = m + 1 ↔ m = 0;%oz: ⊦ m + 1 ≠ 0;%pred: ⊦ (m + 1)-1 = m;%ref: ⊦ ¬ n < n;