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Proof step 2 from proof fst-ls (previous step)

m < m.1 ⊦ m.1 = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ m.1 < m ↔ m ≠ 0 ∧ m ≠ 1

Rule argument

spec: nat-pair inst: name: fst-ls-01 seq: ⊦ m.1 < m ↔ m ≠ 0 ∧ m ≠ 1 subst: [m] → [m] flags: current, []

Additional information

Used simplifier rules: fst-zero: ⊦ 0.1 = 0;%fst-one: ⊦ 1.1 = 1;%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: m