Metamath Proof Explorer


Theorem uzindd

Description: Induction on the upper integers that start at M . The first four hypotheses give us the substitution instances we need; the following two are the basis and the induction step, a deduction version. (Contributed by metakunt, 8-Jun-2024)

Ref Expression
Hypotheses uzindd.1 ( 𝑗 = 𝑀 → ( 𝜓𝜒 ) )
uzindd.2 ( 𝑗 = 𝑘 → ( 𝜓𝜃 ) )
uzindd.3 ( 𝑗 = ( 𝑘 + 1 ) → ( 𝜓𝜏 ) )
uzindd.4 ( 𝑗 = 𝑁 → ( 𝜓𝜂 ) )
uzindd.5 ( 𝜑𝜒 )
uzindd.6 ( ( 𝜑𝜃 ∧ ( 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ) → 𝜏 )
uzindd.7 ( 𝜑𝑀 ∈ ℤ )
uzindd.8 ( 𝜑𝑁 ∈ ℤ )
uzindd.9 ( 𝜑𝑀𝑁 )
Assertion uzindd ( 𝜑𝜂 )

Proof

Step Hyp Ref Expression
1 uzindd.1 ( 𝑗 = 𝑀 → ( 𝜓𝜒 ) )
2 uzindd.2 ( 𝑗 = 𝑘 → ( 𝜓𝜃 ) )
3 uzindd.3 ( 𝑗 = ( 𝑘 + 1 ) → ( 𝜓𝜏 ) )
4 uzindd.4 ( 𝑗 = 𝑁 → ( 𝜓𝜂 ) )
5 uzindd.5 ( 𝜑𝜒 )
6 uzindd.6 ( ( 𝜑𝜃 ∧ ( 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ) → 𝜏 )
7 uzindd.7 ( 𝜑𝑀 ∈ ℤ )
8 uzindd.8 ( 𝜑𝑁 ∈ ℤ )
9 uzindd.9 ( 𝜑𝑀𝑁 )
10 7 8 9 3jca ( 𝜑 → ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀𝑁 ) )
11 1 imbi2d ( 𝑗 = 𝑀 → ( ( 𝜑𝜓 ) ↔ ( 𝜑𝜒 ) ) )
12 2 imbi2d ( 𝑗 = 𝑘 → ( ( 𝜑𝜓 ) ↔ ( 𝜑𝜃 ) ) )
13 3 imbi2d ( 𝑗 = ( 𝑘 + 1 ) → ( ( 𝜑𝜓 ) ↔ ( 𝜑𝜏 ) ) )
14 4 imbi2d ( 𝑗 = 𝑁 → ( ( 𝜑𝜓 ) ↔ ( 𝜑𝜂 ) ) )
15 5 adantr ( ( 𝜑𝑀 ∈ ℤ ) → 𝜒 )
16 15 expcom ( 𝑀 ∈ ℤ → ( 𝜑𝜒 ) )
17 3anass ( ( 𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ↔ ( 𝑀 ∈ ℤ ∧ ( 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ) )
18 ancom ( ( 𝑀 ∈ ℤ ∧ ( 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ) ↔ ( ( 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ∧ 𝑀 ∈ ℤ ) )
19 17 18 bitri ( ( 𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ↔ ( ( 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ∧ 𝑀 ∈ ℤ ) )
20 6 ad4ant123 ( ( ( ( 𝜑𝜃 ) ∧ ( 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ) ∧ 𝑀 ∈ ℤ ) → 𝜏 )
21 20 anasss ( ( ( 𝜑𝜃 ) ∧ ( ( 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ∧ 𝑀 ∈ ℤ ) ) → 𝜏 )
22 19 21 sylan2b ( ( ( 𝜑𝜃 ) ∧ ( 𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ) → 𝜏 )
23 22 3impa ( ( 𝜑𝜃 ∧ ( 𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ) → 𝜏 )
24 23 3com23 ( ( 𝜑 ∧ ( 𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ∧ 𝜃 ) → 𝜏 )
25 24 3expia ( ( 𝜑 ∧ ( 𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) ) → ( 𝜃𝜏 ) )
26 25 expcom ( ( 𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) → ( 𝜑 → ( 𝜃𝜏 ) ) )
27 26 a2d ( ( 𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘 ) → ( ( 𝜑𝜃 ) → ( 𝜑𝜏 ) ) )
28 11 12 13 14 16 27 uzind ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀𝑁 ) → ( 𝜑𝜂 ) )
29 10 28 mpcom ( 𝜑𝜂 )