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Proof step 5 from proof bits2int-nle (previous step)

0 ≤ 2 ^ n→i(# bits - 1) + bits2int(bits), 2 ^ n→i(n) + bits2int(bits) ≤ 0, # bits ≤ n, bits2int(bits) < 2 ^ n→i(# bits - 1), 2 ^ n→i(n) ≤ 2 ^ n→i(# bits - 1) ⊦ bits = []

User applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
# bits = m + 1 ∧ # bits ≠ 0, 0 ≤ 2 ^ n→i(m) + bits2int(bits), 2 ^ n→i(n) + bits2int(bits) ≤ 0, # bits ≤ n, bits2int(bits) < 2 ^ n→i(m), 2 ^ n→i(n) ≤ 2 ^ n→i(m) ⊦ bits = []

Rule argument

seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# bits, m] rotate: T spec: nat inst: name: elim-pred-c

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # bits, m