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Proof step 2 from proof del1 (previous step)
| a ∈ x ⊦ #oc(a, x -1l a) = #oc(a, x) - 1 |
User applied rule insert elim lemma
And obtained the following new goals
- Proof step 3:
#oc(a, x) = m + 1 ∧ #oc(a, x) ≠ 0, a ∈ x ⊦ #oc(a, x -1l a) = m
Rule argument
spec: nat
inst:
name: elim-pred-c
seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1
subst: [n, m] → [#oc(a, x), m]
flags: Additional information
Used simplifier rules:
in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim)
Used terms: #oc(a, x), m